# Is there a physical slowing of time that is only caused by the speed? Simplex0
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### Thank you Odysseus and

Thank you Odysseus and Devilogic for making the world of science a little more understandable. Devilogic
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### Sorry for not writing back

Message 75562 in response to message 75560

Sorry for not writing back sooner, but I have a few comments:

Quote:
Quote:
Will you and the observer in the rocket measure different time between 2 pulses?

No, as long as the sourceâ€™s radial velocity is the same WRT each of us.

Actually, unless the star is moving with the same velocity (not just the radial component) in both frames you will measure a different time between 2 consecutive pulses, because of the different amount of the slowing of time on the star in the reference frame of the Earth and the rocket (and it doesn't matter how far away the star is at all - ie. how many [or little] degrees it traces out on the sky in a given time). Such motion of the star would be quite strange I suppose (I will try to derive it though - seems like a fun little problem).

Just a reminder: By saying that the radial component is the only thing that matters you are ignoring the actual slowing of time (for this the magnitude of the whole velocity is important), and by saying that the radial component is unchanged between both frames you are ignoring the relativistic law of velocity composition (the radial component changes!) - in short, you are ignoring relativity when the speeds involved are 0.9c :)

Best regards Simplex0
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### RE: Sorry for not writing

Message 75563 in response to message 75562

Quote:

Sorry for not writing back sooner, but I have a few comments:

Quote:
Quote:
Will you and the observer in the rocket measure different time between 2 pulses?

No, as long as the sourceâ€™s radial velocity is the same WRT each of us.

Actually, unless the star is moving with the same velocity (not just the radial component) in both frames you will measure a different time between 2 consecutive pulse

Thank you Devilogic!
I must remind my self to specify exactly how each measurement is done.

Assume that you stand on Earth and that the Earth is on the center point of an 2 light year long strait line measured from Earth, 1 light year in each direction, and a rocket with an on board observer located in each end of the line.

Perpendicular from that line and 1 billion light year from Earth there is a variable star that an observer on Earth and the observers in the rockets measure the elapsed time between 2 pulses on. An observer located at the star will measure a relative velocity between each rocket an Earth to 0.9 c towards the Earth.

The observer on Earth measure an elapsed time of 1 second between 2 pulses from the star. How long elapsed time between 2 pulses will the observers in the rockets measure?

What will the relative velocity between Earth and a rocket be measured by the observer on Earth?

What will the relative velocity between Earth and a rocket be measured by the observer on a rocket? Devilogic
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### RE: Thank you Devilogic! I

Message 75564 in response to message 75563

Quote:

Thank you Devilogic!
I must remind my self to specify exactly how each measurement is done.

Assume that you stand on Earth and that the Earth is on the center point of an 2 light year long strait line measured from Earth, 1 light year in each direction, and a rocket with an on board observer located in each end of the line.

Perpendicular from that line and 1 billion light year from Earth there is a variable star that an observer on Earth and the observers in the rockets measure the elapsed time between 2 pulses on. An observer located at the star will measure a relative velocity between each rocket an Earth to 0.9 c towards the Earth.

The observer on Earth measure an elapsed time of 1 second between 2 pulses from the star. How long elapsed time between 2 pulses will the observers in the rockets measure?

What will the relative velocity between Earth and a rocket be measured by the observer on Earth?

What will the relative velocity between Earth and a rocket be measured by the observer on a rocket?

Thanks for the explanation, though there is one piece of information that is still missing: is the star stationary with respect to Earth (easier to calculate things), or is it moving away from Earth (more realistic), and if so at what speed?

I will for simplicity's sake assume that the star and Earth at rest with respect to each other (if that is not what you meant say so). Then the star's inertial frame is also Earth's inertial frame (up to a choice of origin). This means that the velocity between Earth and a rocket is the same measured from Earth as it is measured from the star (that is: v0 = 0.9c).

Now, we have to switch frames to see what the observer on a rocket sees. First we are interested in velocities. I will mark star's/Earth's inertial frame as S and the rocket's frame as S'. I will mark this rocket (the rocket that is at rest in S') as "rocket A" and the other one as "rocket B". In S we have the following velocities (written out by components for pedagogical purposes):

- rocket A, frame S': vA = ( v0, 0)
- rocket B: vB = (-v0, 0)
- Earth, star, frame S: vE = (0, 0)

Now, the relativistic transformation law for velocities is the following (vX marks [signed!] velocity [in frame S] in the direction in which frame S' is moving, and vY velocity in the perpendicular direction, vX' and vY' are the transformed velocities [that is velocities in S'], and Gamma is the relativistic gamma: Gamma=1/Sqrt(1-(v0/c)^2) ):

vX' = (vX-v0)/(1-vX*v0/c^2)
vY' = vY/(Gamma*(1-vX*v0/c^2))

If we apply these formulae to our problem we arrive at the following velocities in S':

- rocket A, frame S': vA' = (0, 0)
- rocket B: vB' = (-2*v0/(1+(v0/c)^2), 0) ~= (-0.9945c, 0)
- Earth, star, frame S: vE' = (-v0, 0)

This means: from the rocket A's POV (both rockets are equivalent in this experiment though, since they have symmetrical velocities with regards to the Earth and the star) Earth is moving towards it with velocity v0=0.9c and rocket B is moving towards it with velocity ~0.9954c.

---

Next up, time between two pulses. An event (t,x,y) in S has the following transformation to (t',x',y') in S' in special relativity:

t' = Gamma*(t - v0/c^2*x)
x' = Gamma*(x - v0*t)
y' = y

What we are looking at with regards to time between two pulses is the following two events in the inertial frame of the star S (you can also pick those two events in Earth's frame, but that is in general harder to do - but since Earth happens to be at rest in frame S it makes no difference in our case): E1=(0,0,0) and E2=(t0,0,0), where t0 is the time between two pulses from the star's POV. (Here I set the origin of frame S on the star - not that it matters :). Since Earth is at rest in S the time between two pulses from Earth's POV is also t0, and therefore t0=1s.

In the rocket's inertial frame S' events E1 and E2 transform to:

E1' = (0, 0, 0)
E2' = (Gamma*t0, -Gamma*v0*t0, 0)

This means that we can say that the time between two pulses from the star is (physically!): t' = E2't - E1't = Gamma*t0 ~= 2,29s in the rocket's frame.

These have all been effects of the "2nd kind" (physically real effects, not just observational effects). Now we have to take stock of effects of the "1st kind" (observational effects because of the slowness of light). Since the star is sufficiently far away from the rocket we can ignore the change in the "x" position of the star and say that pulses from the star are all arriving approximately in the perpendicular direction to the direction of the rocket ((from Earth's POV)).

If the star was moving in the "y" direction we would have to add the following to the time between two pulses (vSY' is the velocity of the star in the y direction in frame S' [in our case vSY'=0]): (E2'y-E1'y)/c = t'*(vSY'/c) because light has to travel the extra distance between the places of emission of both pulses. This means that the observed time between two pulses would be:

t'' = t'*(1 + vSY'/c)

Since in our case vSY'=0 this implies that t'' = t', and the observed time between two pulses is the same as the physical time between two pulses (=> in this experiment effects of the 1. kind have no effect :).

Hope this helps ;)

P.S.
If the star were not stationary with respect to Earth you would have to take into account it's full velocity in the calculations (both the x and the y component, which would BOTH change when you would switch frames from Earth to the rocket). If you need me to I can show you how to do that as well, but it shouldn't be too hard to figure it out on your own :) (just remember that you now have 3 relevant inertial frames - the star's, the Earth's and the rocket's). Simplex0
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### RE: RE: Thank you

Message 75565 in response to message 75564

Quote:
Quote:

Thank you Devilogic!
I must remind my self to specify exactly how each measurement is done.

Assume that you stand on Earth and that the Earth is on the center point of an 2 light year long strait line measured from Earth, 1 light year in each direction, and a rocket with an on board observer located in each end of the line.

Perpendicular from that line and 1 billion light year from Earth there is a variable star that an observer on Earth and the observers in the rockets measure the elapsed time between 2 pulses on. An observer located at the star will measure a relative velocity between each rocket an Earth to 0.9 c towards the Earth.

The observer on Earth measure an elapsed time of 1 second between 2 pulses from the star. How long elapsed time between 2 pulses will the observers in the rockets measure?

What will the relative velocity between Earth and a rocket be measured by the observer on Earth?

What will the relative velocity between Earth and a rocket be measured by the observer on a rocket?

Thanks for the explanation, though there is one piece of information that is still missing: is the star stationary with respect to Earth

Sorry I missed that. Yes the star is stationary with respect to Earth. Simplex0
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### Thank you Devilogic for all

Thank you Devilogic for all the help. Devilogic
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### Glad to have been of

Message 75567 in response to message 75566

Glad to have been of assistance :) Czar Brent
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