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 Post subject: Posted: Wed Feb 08, 2012 2:08 pm

Joined: Wed Jul 14, 2010 5:27 pm
Posts: 1011
tomclarke wrote:
charliem wrote:
tomclarke wrote:
Thus the S1/S2 clock time is identical to the previous B time, and shorter than the OB time. However you cannot calculate the elapsed time of OB clock from S1/S2 without a time shift correction, just the ame as in the B case.

Sorry Tom, I don't follow you there.

Do you mean that when they compare OB's and S2's clocks they say the same time, or that S2's clock is behind?

S2 is behind.

Look at it from OB's frame, the S1/S2 situation is the same as the B situation in my A/B example, and the S1/S2 clocks are behind.

Look at it from S1/S2 frames (which is why you introduced S1/S2). The two journeys are as before, with OB clock slightly behind S1 and also S2. However in changing from S1 frame to S2 frame the same time shift happens as before with B, which makes S1/S2 time slightly behind OB time.

This frame change is only an issue when calculating OB time that corresponds to S1 or S2 time using LT. It is what turns a symmetrical calculation (the two journeys) into an asymmetrical one.

yes, it's the fact that it's tomclarke special inertial reference frame, a.k.a. "The Chosen Frame".

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 Post subject: Posted: Wed Feb 08, 2012 5:35 pm

Joined: Thu Jan 12, 2012 11:18 am
Posts: 93
im no longer sure how the universe is functioning at all with all this going on

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 Post subject: Posted: Thu Feb 09, 2012 6:18 pm

Joined: Sun Oct 05, 2008 4:52 pm
Posts: 1452
Location: London
happyjack27 wrote:
tomclarke wrote:
charliem wrote:
tomclarke wrote:
Thus the S1/S2 clock time is identical to the previous B time, and shorter than the OB time. However you cannot calculate the elapsed time of OB clock from S1/S2 without a time shift correction, just the ame as in the B case.

Sorry Tom, I don't follow you there.

Do you mean that when they compare OB's and S2's clocks they say the same time, or that S2's clock is behind?

S2 is behind.

Look at it from OB's frame, the S1/S2 situation is the same as the B situation in my A/B example, and the S1/S2 clocks are behind.

Look at it from S1/S2 frames (which is why you introduced S1/S2). The two journeys are as before, with OB clock slightly behind S1 and also S2. However in changing from S1 frame to S2 frame the same time shift happens as before with B, which makes S1/S2 time slightly behind OB time.

This frame change is only an issue when calculating OB time that corresponds to S1 or S2 time using LT. It is what turns a symmetrical calculation (the two journeys) into an asymmetrical one.

yes, it's the fact that it's tomclarke special inertial reference frame, a.k.a. "The Chosen Frame".

You have to choose to do the calculations

I explained how you can do the calculation in different frames, and still get the same answer. But there is only one easy inertial frame to choose, since B (or S1/S2) has two frames not one, which makes things much more complex. Still, you will get the same answer whatever frame you choose. Try it!

Last edited by tomclarke on Fri Feb 10, 2012 7:33 pm, edited 1 time in total.

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 Post subject: Posted: Thu Feb 09, 2012 7:22 pm

Joined: Thu Jan 12, 2012 11:18 am
Posts: 93
sry newbie question but why does clock A run at t/gamma, not t , if it is stationary

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 Post subject: Posted: Thu Feb 09, 2012 11:56 pm

Joined: Wed Jul 14, 2010 5:27 pm
Posts: 1011
tomclarke wrote:
happyjack27 wrote:
tomclarke wrote:
charliem wrote:
tomclarke wrote:
Thus the S1/S2 clock time is identical to the previous B time, and shorter than the OB time. However you cannot calculate the elapsed time of OB clock from S1/S2 without a time shift correction, just the ame as in the B case.

Sorry Tom, I don't follow you there.

Do you mean that when they compare OB's and S2's clocks they say the same time, or that S2's clock is behind?

S2 is behind.

Look at it from OB's frame, the S1/S2 situation is the same as the B situation in my A/B example, and the S1/S2 clocks are behind.

Look at it from S1/S2 frames (which is why you introduced S1/S2). The two journeys are as before, with OB clock slightly behind S1 and also S2. However in changing from S1 frame to S2 frame the same time shift happens as before with B, which makes S1/S2 time slightly behind OB time.

This frame change is only an issue when calculating OB time that corresponds to S1 or S2 time using LT. It is what turns a symmetrical calculation (the two journeys) into an asymmetrical one.

yes, it's the fact that it's tomclarke special inertial reference frame, a.k.a. "The Chosen Frame".

You have to choose to do the calculations

Iexplained how you can do the calculation in different frames, and still get the same answer. But there is only one easy inertial frame to choose, since B (or S1/S2) has two frames not one, which makes things much more complex. Still, you will get the same answer whatever frame you choose. Try it!

"... you will get the same answer whatever frame you choose. "

OMG it's a miracle, you finally understand!
t(b in a) < t(a in a)
t(a in b) < t(b in b)
t(b in a) - t(a in a) = t(a in b) - t(b in b)

ergo, if
t(a in a) = t(b in b)
then
t(a in b) = t(b in a)

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 Post subject: Posted: Fri Feb 10, 2012 7:08 am

Joined: Sun Oct 05, 2008 4:52 pm
Posts: 1452
Location: London
happyjack27 wrote:
tomclarke wrote:
happyjack27 wrote:
tomclarke wrote:
charliem wrote:
tomclarke wrote:
Thus the S1/S2 clock time is identical to the previous B time, and shorter than the OB time. However you cannot calculate the elapsed time of OB clock from S1/S2 without a time shift correction, just the ame as in the B case.

Sorry Tom, I don't follow you there.

Do you mean that when they compare OB's and S2's clocks they say the same time, or that S2's clock is behind?

S2 is behind.

Look at it from OB's frame, the S1/S2 situation is the same as the B situation in my A/B example, and the S1/S2 clocks are behind.

Look at it from S1/S2 frames (which is why you introduced S1/S2). The two journeys are as before, with OB clock slightly behind S1 and also S2. However in changing from S1 frame to S2 frame the same time shift happens as before with B, which makes S1/S2 time slightly behind OB time.

This frame change is only an issue when calculating OB time that corresponds to S1 or S2 time using LT. It is what turns a symmetrical calculation (the two journeys) into an asymmetrical one.

yes, it's the fact that it's tomclarke special inertial reference frame, a.k.a. "The Chosen Frame".

You have to choose to do the calculations

Iexplained how you can do the calculation in different frames, and still get the same answer. But there is only one easy inertial frame to choose, since B (or S1/S2) has two frames not one, which makes things much more complex. Still, you will get the same answer whatever frame you choose. Try it!

"... you will get the same answer whatever frame you choose. "

OMG it's a miracle, you finally understand!
t(b in a) < t(a in a)
t(a in b) < t(b in b)
t(b in a) - t(a in a) = t(a in b) - t(b in b)

ergo, if
t(a in a) = t(b in b)
then
t(a in b) = t(b in a)

That is all true, but it applies only to the two single frame parts of the journey, not to the comparison of clocks when the two twins are again together.

That is because t(a in b over journey) = t(a in b) _S! + t(a in b)S2 + t(a change from S1 to S2)

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 Post subject: Posted: Fri Feb 10, 2012 7:09 am

Joined: Sun Oct 05, 2008 4:52 pm
Posts: 1452
Location: London
303 wrote:
sry newbie question but why does clock A run at t/gamma, not t , if it is stationary

t/gamma is rate A clock runs in B's frame. All time is relative to frame.

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 Post subject: Posted: Fri Feb 10, 2012 1:49 pm

Joined: Wed Jul 14, 2010 5:27 pm
Posts: 1011
tomclarke wrote:
happyjack27 wrote:
tomclarke wrote:
happyjack27 wrote:
tomclarke wrote:
charliem wrote:
tomclarke wrote:
Thus the S1/S2 clock time is identical to the previous B time, and shorter than the OB time. However you cannot calculate the elapsed time of OB clock from S1/S2 without a time shift correction, just the ame as in the B case.

Sorry Tom, I don't follow you there.

Do you mean that when they compare OB's and S2's clocks they say the same time, or that S2's clock is behind?

S2 is behind.

Look at it from OB's frame, the S1/S2 situation is the same as the B situation in my A/B example, and the S1/S2 clocks are behind.

Look at it from S1/S2 frames (which is why you introduced S1/S2). The two journeys are as before, with OB clock slightly behind S1 and also S2. However in changing from S1 frame to S2 frame the same time shift happens as before with B, which makes S1/S2 time slightly behind OB time.

This frame change is only an issue when calculating OB time that corresponds to S1 or S2 time using LT. It is what turns a symmetrical calculation (the two journeys) into an asymmetrical one.

yes, it's the fact that it's tomclarke special inertial reference frame, a.k.a. "The Chosen Frame".

You have to choose to do the calculations

Iexplained how you can do the calculation in different frames, and still get the same answer. But there is only one easy inertial frame to choose, since B (or S1/S2) has two frames not one, which makes things much more complex. Still, you will get the same answer whatever frame you choose. Try it!

"... you will get the same answer whatever frame you choose. "

OMG it's a miracle, you finally understand!
t(b in a) < t(a in a)
t(a in b) < t(b in b)
t(b in a) - t(a in a) = t(a in b) - t(b in b)

ergo, if
t(a in a) = t(b in b)
then
t(a in b) = t(b in a)

That is all true, but it applies only to the two single frame parts of the journey, not to the comparison of clocks when the two twins are again together.

That is because t(a in b over journey) = t(a in b) _S! + t(a in b)S2 + t(a change from S1 to S2)

utterly pointless. you still don't frickin' get it. you can put whatever you want in there.

if

t(a in b over journey) = t(a in b) _S! + t(a in b)S2 + t(a change from S1 to S2)

then also

t(b in a over journey) = t(b in a) _S! + t(b in a)S2 + t(b change from S2 to S1)

duh.

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 Post subject: Posted: Fri Feb 10, 2012 7:22 pm

Joined: Sun Oct 05, 2008 4:52 pm
Posts: 1452
Location: London
happyjack wrote:
utterly pointless. you still don't frickin' get it. you can put whatever you want in there.

if

t(a in b over journey) = t(a in b) _S! + t(a in b)S2 + t(a change from S1 to S2)

then also

t(b in a over journey) = t(b in a) _S! + t(b in a)S2 + t(b change from S2 to S1)

duh.

I think we are finally getting somewhere.

There is no change in the A frame, since A stays in one inertial frame. There is a change in the B frame, from S1 to S2.

Hence the two equations are not the same, viewed from B, which changes frame, or A, which does not.

I have been saying this for some time, but perhaps this will help you more than previous posts.

Note that the time shift element is nothing to do with relative velocity changing. When calculated in B's frames it is independent of the velocity of A, and depends only on the change in velocity (and therefore frame) of B and the distance between B and A in the S1 (or S2) frame. Thus it is not symmetrical between A - which never changes frame, and B, which changes from S1 to S2.

Time shift is a correction needed in a relativistic world when an observer's frame changes. It affects all his calculated times of distant clocks. These times must be calculated because direct comparison is impossible. The change in (calculated) time of a distant clock seems a bit weird but not when you consider that in relativistic space there is no such thing as absolute "now". The set of spacetime events corresponding to "now" depends on the frame in which "now" is measured.

Thus the distant clocks do not change time, rather the distant clock time considered the same as the local time varies when the local frame changes.

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 Post subject: Posted: Fri Feb 10, 2012 8:18 pm

Joined: Thu Jan 12, 2012 11:18 am
Posts: 93
so the twins are the same age then, distant time remains normal, and only the calculated time is off ?

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 Post subject: Posted: Sat Feb 11, 2012 10:03 am

Joined: Sun Oct 05, 2008 4:52 pm
Posts: 1452
Location: London
303 wrote:
so the twins are the same age then, distant time remains normal, and only the calculated time is off ?

Calculated time becomes measurable when two clocks are brought together again. It is calculated in the sense that when the clocks are distant there times cannot be directly conpared, but can be calculated in any given reference frame.

There is no absolute standard of time. You can compare the two clock times in whatever frame you like, and track calculated time from start to end of the journey to see what is the difference. When the clocks come together again the time difference can be measured and will be the same in all frames.

The Lorentz transformation tells you that B's clock, in A's frame, consistently runs slow relative to A's clock. Thus at the end of the journey B will be younger than A when they meet.

This whole thread is about people (Johan, Happyjack) having a misconceived argument that this can't be true. The two underlying ideas are:

(1) "clocks keep the same time, so must read the same when brought together again. Any LT effect must be "apparent" not "real".

That is misunderstanding time in a relativistic world. Two different trajectories through spacetime can have different clock times even though they meet. Of course, our intuition tells that that is not possible because there is only one "now", and global time.

(2) The relative velocity and so LT time dilation is symmetrical. Therefore it cannot result in a time difference.

That is the subtlest argument, and in the last few posts we have just seen why it is not true.

The point is that to see which clock will go fastest we can only calculate the time of one (distant) clock relative to the (instantaneous) frame of the other (local). When the clocks are brought together again the calculated time can be read directly. When the clocks are far away calculation is all we have.

So, on the two journeys each clock runs faster than the calculated time of the other. It is symmetrical. However the B clock has to change frame to come home, and this change of frame alters the A calculated time ahead by double the amount the Lorentz Transformation slowed it down, so that at the end the time difference calculation in B's two rest frames is identical to that in A's single rest frame.

Thus the end result is independent of frame: A's clock has a longer elapsed time than B's clock over the journey.

You need not think of this as clocks running faster or slower. You can think of it as time depending on spacetime path: bent paths have shorter times. Actually this is intuitive. The most highly possible bent path (with highest relative velocity to an inertial reference frame) would be a ray of light travelling out and reflected back. The time elapsed along this ligt-ray path is known to be zero.

Your intuition will make mistakes because you are naturally wanting to compare times in different frames, and expect the results to be independent of frame. But there is no unique way to compare times of distant clocks in different frames. So the LT can be correct, even though the results it gives are not compatible with a unique global comparison of time.

The LT apprent contradiction in calculated times is possible because two clocks in different frames can only ever meet once, so the apparent discrepancy is resolved by the time shift difference in calculated time of distant objects.

When one clock returns the elapsed time difference can be precisely measured. The LT calculated times can be checked, since when two clocks are together they can be directly compared. There is never inconsistency because in order to return one at least of the clocks must change frame and therefore add an asymmetrical tiume shift component to the calculated elapsed time.

If the two trajectories are symmetrical, with both A and b clocks voyaging out and returning, with the same delta V change at the apex of the voyage, the two clocks would of course end up reading the same.

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 Post subject: Posted: Sat Feb 11, 2012 5:06 pm

Joined: Wed Jul 14, 2010 5:27 pm
Posts: 1011
tomclarke wrote:
happyjack wrote:
utterly pointless. you still don't frickin' get it. you can put whatever you want in there.

if

t(a in b over journey) = t(a in b) _S! + t(a in b)S2 + t(a change from S1 to S2)

then also

t(b in a over journey) = t(b in a) _S! + t(b in a)S2 + t(b change from S2 to S1)

duh.

I think we are finally getting somewhere.

There is no change in the A frame, since A stays in one inertial frame. There is a change in the B frame, from S1 to S2.

Hence the two equations are not the same, viewed from B, which changes frame, or A, which does not.

I have been saying this for some time, but perhaps this will help you more than previous posts.

Note that the time shift element is nothing to do with relative velocity changing. When calculated in B's frames it is independent of the velocity of A, and depends only on the change in velocity (and therefore frame) of B and the distance between B and A in the S1 (or S2) frame. Thus it is not symmetrical between A - which never changes frame, and B, which changes from S1 to S2.

Time shift is a correction needed in a relativistic world when an observer's frame changes. It affects all his calculated times of distant clocks. These times must be calculated because direct comparison is impossible. The change in (calculated) time of a distant clock seems a bit weird but not when you consider that in relativistic space there is no such thing as absolute "now". The set of spacetime events corresponding to "now" depends on the frame in which "now" is measured.

Thus the distant clocks do not change time, rather the distant clock time considered the same as the local time varies when the local frame changes.

we are not changing frames we are comparing them. if you want to introduce a frame change then you need to also introduce a frame change in the other side, otherwise you're talking about two different trajectories. - comparing apples to oranges.

i see now you are assuming the conclusion. you are comparing apples to oranges and wo and behold they don't equate. i am comparing apples to apples. have you not figured that out yet?

Last edited by happyjack27 on Sat Feb 11, 2012 5:10 pm, edited 1 time in total.

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 Post subject: Posted: Sat Feb 11, 2012 5:09 pm

Joined: Wed Jul 14, 2010 5:27 pm
Posts: 1011
303 wrote:
so the twins are the same age then, distant time remains normal, and only the calculated time is off ?

twin a thinks he's x years older than twin b, and seemingly paradoxically (hence twin paradox) twin b also thinks he's x years older than twin a.

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 Post subject: Posted: Sat Feb 11, 2012 5:34 pm

Joined: Thu Jan 12, 2012 11:18 am
Posts: 93
tx for replies, i think i'm least comfortable with the concept of time as a integrated part of spacial volume, spacetime. i always thought of it as just an index for 3d coords

so whats actually changing for one, or each!, of the twins, their rate of elapsed time is slowed ? this is what is meant by time dilation

is anyone sure this isn't some maths quirk, quantisation issue, overflow or something like that?

im curious, remember as a non mathematician, how LT breaks down to the gallilean function, whats the speed cutoff point for this, some point at which you'd be better off using the old fashioned way , can LT be used right down to walking pace and still give the right results?

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 Post subject: Posted: Sat Feb 11, 2012 5:55 pm

Joined: Sun Oct 05, 2008 4:52 pm
Posts: 1452
Location: London
303 wrote:
tx for replies, i think i'm least comfortable with the concept of time as a integrated part of spacial volume, spacetime. i always thought of it as just an index for 3d coords

so whats actually changing for one, or each!, of the twins, their rate of elapsed time is slowed ? this is what is meant by time dilation

is anyone sure this isn't some maths quirk, quantisation issue, overflow or something like that?

im curious, remember as a non mathematician, how LT breaks down to the gallilean function, whats the speed cutoff point for this, some point at which you'd be better off using the old fashioned way , can LT be used right down to walking pace and still give the right results?

Yes, LT is it, and at slow speeds the differences are small but still present. The factor is sqrt(1-v^2/c^2) ~ 0.5*v^2/c^2 as a fractional correction.

The "simple" headline is that the moving twin undergoes time dilation by this factor. This actually gives the right answer. But it hides the complexity of what time dilation really is, because of course it is because the "moving" twin goes and comes back, with frame change in middle, that he is considered moving. Given all motion is relative the thing which leads to his time "genuinely" being dilated is that his motion changes - so there is no inertial frame in which he is always at rest.

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