# Time dilation: speed relative to what?

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1. Mar 19, 2015

### Grimble

I have been asked about time dilation which I thought I understood, but maybe you can help me provide an answer?
If a spaceship is travelling at 0.866c then time will slow by a factor of 2 due to the speed the spaceship has.

Why?

If it is alone in space, light years away from any other body, how can that speed affect the clock?
What is the speed talked of so glibly actually relative to?

If two spaceships were travelling in opposite directions at 0.433c (lorentz factor = 1.109) then for an observer midway between their clocks would each be running slow at 0.901 x a standard clock, i.e. keeping the same time, yet each spaceship would measure the other's time as being slow by a factor of 2.

So how can one say the clocks are actually running slow, rather than just being measured to run slow by the moving observers?

2. Mar 19, 2015

### A.T.

What is the difference between "measured to run slow" and "actually running slow"?

3. Mar 19, 2015

### Staff: Mentor

Whomever told you what the speed is will have to also tell you what it is relative to. Yes, speed only has meaning when you specify the frame of reference from which it is measured.

For the other: have you read-up on the twins paradox? You've proposed a totally symmetrical scenario -- most are not. The symetry or lack thereof matters for who is the one who's clock shows the less elapsed time.

4. Mar 19, 2015

### Grimble

The difference between 'measured to run slow' and 'actually run slow' is that, according to the other spaceship, the clock would be running at half the speed of the one spaceship, while both clocks in the spaceships are running, not at 0.5 but at 0.901 of a standard clock. (Which of course, to comply with Einstein's postulate would run at the same rate in any inertial Frame of Reference). And, as measured by the midway observer both clocks are keeping identical time.

The question specified a lone spaceship, far from any other body, travelling at 0.866c...
I would take it that the implication is that it is relative to space, no other FoR was specified. Am I to tell the questioner that they have to Specify the FoR in which that is measured? Difficult to justify doing that when we are told that time slows when nearing the speed of light, and we have to specify relative to some arbitrary observer...

As for symmetry, there is symmetry. This is not the twin paradox where one has accelerated away from the other, these are two spaceships, each travelling travelling at 0.433c in opposite directions away from a central observer. No one has accelerated, they are inertial Frames of Reference.

Therefore the clocks must be running at the same rate as measured by that central observer - 0.901 of the standard clock held by the central observer who is at rest in space and whose clock must therefore be measuring proper time.

So is 0.901 proper time the true rate of the spaceships clocks? And if so how do they each measure the other's clock to be running at half the rate of their own clock?
(And how many different times can a singe clock measure?)

5. Mar 19, 2015

### phinds

Clocks always run at one second per second unless they are broken. BOTH spaceships see their clocks running at one second per second and the other spaceship's clock running slower.

6. Mar 19, 2015

### Grimble

So, what you are saying is that both clocks are running normally - as the first postulate requires, but when measuring a moving clock, i.e. measuring under different conditions that clock is measured to run slower. Is that not the answer - under those conditions the measurement of the moving clock is affected by the relative speed?
The question is then, how? What is the mechanism by which that measurement is affected...

It cannot be that the clock actually runs slow for their is no connection between the clock and the observer for whom it is in motion.
I.e. if the 'stationary' observer changed his speed there is no connection to the 'moving' clock to promote that change in RELATIVE velocity onto the clock, no way in which the rate of that clock could physically change relative to the observer.
It can only be that the MEASUREMENT BY the remote observer is affected.
The clocks don't change only the measurements? That can't be right surely????

7. Mar 19, 2015

### Khashishi

The gist of relativity is that the view of every observer (in an inertial reference frame) is equally valid. In relativity, there is no "standard" clock.

8. Mar 19, 2015

### Bandersnatch

9. Mar 19, 2015

### phinds

In case you don't realize it, you, right now as you read this, are moving at .99999c. You are also moving at .8c You are also moving at .1 c. You are also not moving at all. In each of these frames, your clock appears different, so of course, as others have already pointed out, it IS the measurement that changes.

10. Mar 19, 2015

### Ibix

Both clocks running slow is actually quite closely analogous to the following.

Two cars are doing 30mph side-by-side down a road. They each appear stationary to each other.

One car turns off onto a road that runs at an angle $\theta$ to the first road. That car starts to fall behind according to the one continuing straight on, because its velocity in the straight-on direction is $30\cos\theta$. However, the other car can say the exact same thing and for the same reason. So both cars are falling behind according to the other.

This happens because the cars no longer agree on what "forward" is. In the case of moving spaceships (and the cars, although their velocity is so low you'll never notice), the two no longer agree what is "forward" through spacetime so they can both say the other is moving more slowly through time.

Health warning: this is only an analogy. It's a good one, but don't go crazy with it.

11. Mar 19, 2015

### A.T.

Remoteness has nothing to do with it. Time dilation depends on relative motion, not on distance. And in physics there is no difference between what is measured, and what is "actually". Physics cares only about what is measured.

12. Mar 19, 2015

### Rising Eagle

I'm with the original poster. A photon has no frame of reference needed. It always has no rate of ticking of its internal clock no matter what, even if there are no observers of any kind. This almost implies some kind of universal frame of reference. SR only deals with observers at lower than light speed and how they perceive the rate of clock ticking of other travelers' internal clocks when the observers see those other travelers. But a single lonely traveler at lower than light speed will have some nonzero rate of ticking of the clock. This cannot be measured other than with respect to some universal frame like the CMBR.

13. Mar 19, 2015

### Staff: Mentor

You do realize that the OP said nothing at all about photons, right?

No, it doesn't.

Incorrect. SR can describe light just fine. Light can never be at rest in any inertial frame, but that doesn't mean it can't be described using inertial frames.

Incorrect. You can measure clock rates with respect to any inertial frame. And the "frame" provided by the CMBR is not an inertial frame (nor is it a "universal frame"--there is no such thing), because the universe as a whole, including the CMBR and which observers see it as isotropic, is not described by SR, since SR only works when spacetime is flat, and the spacetime of the universe as a whole is curved.

14. Mar 19, 2015

### DaveC426913

I have never heard it described in such a way. That is very clever and very elegant. Especially the 'why' (bolded).

15. Mar 20, 2015

### Grimble

I see that description by Ibix, referred to by DVEC426913, as an illustration of Minkowski Hyperbolic Rotation - if you know what I mean?

@Khashishi: By a standard clock, I mean one keeping Proper Time, that is one that is measuring time for a particle that it is adjacent to. The same Proper Time that is measured by a stationary clock at the origin of a Frame of Reference.

And in physics there is no difference between what is measured, and what is "actually". Physics cares only about what is measured.

So if I measure two people who are 6ft tall but one is further away, and so perspective means that I measure him as only half the height of the other (Let us say they are in floating in space and I view then through the window of my space craft) then those measurements tell me that he IS only half the height of the nearer one?

For that is what you are claiming.

No, the conditions under which a measurement are taken affects that measurement, hence the need for standard conditions, temperatures etc.

As I understand it, taking a measurement from one Inertial Frame of Reference and converting it to make it relative to another Inertial Frame of Reference is done by the Lorentz transformation equations.

So, for example, a proper time in the target iFoR can be converted to a coordinate time in the observer's iFoR.

16. Mar 20, 2015

### A.T.

No, you would measure half angular size, which is different from measuring linear height. Relativistic effects have nothing to do with visual effects, like perspective, signal delay etc. They are what still remains after you have already accounted for these effects.

Last edited: Mar 20, 2015
17. Mar 20, 2015

### stevendaryl

Staff Emeritus
Pick ANY inertial coordinate system. Then a moving clock will experience time dilation relative to that coordinate system. How can it be simultaneously time-dilated according to EVERY single inertial coordinate system? Well, it is.

For simplification, let's just consider 1-dimensional motion in a single direction. Then the time dilation formula is, in infinitesimal form:

$\delta \tau = \sqrt{\delta t^2 - \frac{1}{c^2}\delta x^2}$

meaning that if a clock moves from the point $x$ at time $t$ to the point $x+\delta x$ at time $t+\delta t$, then the time on the clock will advance by an amount $\delta \tau$ as given above. The faster the clock is traveling, the bigger $\delta x$ will be, the smaller $\delta \tau$ will be.

The essential thing to realize about relativity is that even though $\delta t$ is different in different reference frames, and $\delta x$ is different in different reference frames, the combination $\sqrt{\delta t^2 - \frac{1}{c^2}\delta x^2}$ is the same in EVERY reference frame. So every single inertial reference frame agrees that the above time dilation formula works.

So "relative to what" has the answer: "relative to ANY inertial reference frame".

18. Mar 20, 2015

### robphy

Along the lines of what Ibix (above) said,
this is a variant of what I made for my relativity class this semester.
(You can actually pursue the analogy quite far... a paper has been in the works.)

Here is an interactive version (also try the E-slider to see the analogy)
https://www.desmos.com/calculator/ti58l2sair

Here is a physical description of the above
based on an older post (https://www.physicsforums.com/threads/time-dilation.667761/#post-4248601)
in the interactive graph above, use v1=tanh(arccosh(70/50))=0.69985=(approx 0.70) and v2=0 and note that 50/70=0.714.
(Center the intersection point and zoom in. By clicking near the intersection, the graph gives the coordinates.)

Last edited: Mar 20, 2015
19. Mar 20, 2015

### Grimble

20. Mar 20, 2015

### Ibix

You aren't measuring height, though, you are measuring angular size. I think that was AT's point.