Time Dilation

chinglu said:
In step 3 #117, you claimed (x',t') = (-k, k/v) when t'=t=0 with k > 0.
Assume a clock is located at -k in the primed system.
Now, let's shoot a light beam at that clock when the origins are the same.
Hence, while light moves c from the origin, x' moves toward the light at v cutting the distance light must travel.
ct = vt + k/γ
Click to expand...
The clock is at -k, not at k, and the light is moving in the negative direction (is going at -c). So:

$$-ct = vt - k/\gamma$$
$$t = \frac{k}{\gamma(c+v)}$$
$$x = -k/\gamma$$

LT...
$$t' = (t - \frac{vx}{c^2})\gamma$$

$$t' = (\frac{k}{\gamma(c+v)} + \frac{vk}{\gamma c^2})\gamma$$

$$t' = \frac{k(v+1)}{c(v+1)}$$

$$t' = k/c$$

Which is, of course, consistent with light moving at -c from x'=0 at t'=0 to x'=-k at t'=k/c.
 
Pete said:
The clock is at -k, not at k, and the light is moving in the negative direction (is going at -c). So:

$$-ct = vt - k/\gamma$$
$$t = \frac{k}{\gamma(c+v)}$$
$$x = -k/\gamma$$

LT...
$$t' = (t - \frac{vx}{c^2})\gamma$$

$$t' = (\frac{k}{\gamma(c+v)} + \frac{vk}{\gamma c^2})\gamma$$

$$t' = \frac{k(v+1)}{c(v+1)}$$

$$t' = k/c$$

Which is, of course, consistent with light moving at -c from x'=0 at t'=0 to x'=-k at t'=k/c.
Click to expand...

Looks Ok, should have checked my work.

that is because that part is not relevent.

The clock at x, corresponding to x'=-k had t=-γkv/c², when the origins were the same.
Either way, it is a contradiction since you would have applied the wrong t in LT. So, you backed a losing position and find a +- error in my calculations. Thanks for your attention to the detail.
 
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Right, I was hasty and didn't check things through. There's some really dodgy arithmetic in there!
This step is embarrassingly wrong:
Pete said:
$$t' = (\frac{k}{\gamma(c+v)} + \frac{vk}{\gamma c^2})\gamma$$

$$t' = \frac{k(v+1)}{c(v+1)}$$
Click to expand...
So, let's try again, carefully:

chinglu said:
Assume a clock is located at -k in the primed system.
Now, let's shoot a light beam at that clock when the origins are the same.
Hence, while light moves c from the origin, x' moves toward the light at v cutting the distance light must travel.
Click to expand...
We've fixed this bit:
ct = vt + k/γ
t = k/(γ(c-v))
Click to expand...
$$-ct = vt - k/\gamma$$
$$t = \frac{k}{\gamma(c+v)}$$

But there's another mistake here:
x = -k/γ
Click to expand...
The clock at x'=-k was at x=-k/γ when the t clock there had t=0 (i.e. in the unprimed frame, this is when the light beam started.)
But, we just figured out that the light doesn't reach that clock until t = k/γ(c+v)
By that time, the clock at x'=-k has moved from x=-k/γ to:

$$x = -k/\gamma + vt$$

$$x = -k/\gamma + \frac{vk}{\gamma(c+v)}$$

According to LT,
t' = ( t - vx/c² )γ
Plug in the numbers.
Click to expand...
Plugging in the numbers suddenly got a lot harder!
 
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Pete said:
Right, I was hasty and didn't check things through. There's some really dodgy arithmetic in there!
This step is embarrassingly wrong:

So, let's try again, carefully:


We've fixed this bit:

$$-ct = vt - k/\gamma$$
$$t = \frac{k}{\gamma(c+v)}$$

But there's another mistake here:

The clock at x'=-k was at x=-k/γ when the t clock there had t=0 (i.e. in the unprimed frame, this is when the light beam started.)
But, we just figured out that the light doesn't reach that clock until t = k/γ(c+v)
By that time, the clock at x'=-k has moved from x=-k/γ to:

$$x = -k/\gamma + vt$$

$$x = -k/\gamma + \frac{vk}{\gamma(c+v)}$$


Plugging in the numbers suddenly got a lot harder!
Click to expand...

Interesting post.

If you will look at JamesR post #117 #2 in the stationary frame/unprimed frame, you will find an initial condition at x=-k/γ with a spooky at a distance desynchronization on the stationary frame clock which is not valid under SR.

I would be curious of your interpretation of #2 before I proceed.
 
Nothing new here, chinglu, except a bunch of new errors from you. I won't bother replying to your new errors. I'll wait for you to admit your old errors.
 
James R said:
Nothing new here, chinglu, except a bunch of new errors from you. I won't bother replying to your new errors. I'll wait for you to admit your old errors.
Click to expand...

Well, we have figured out in step 2 of the stationary frame of your post #117, you have added a spooky at a distance time to the clock there even though clocks in the stationary frame are assumed to be Einstein synched at t'=t=0.

Can you explain that?
 
chinglu said:
Well, we have figured out in step 2 of the stationary frame of your post #117, you have added a spooky at a distance time to the clock there even though clocks in the stationary frame are assumed to be Einstein synched at t'=t=0.

Can you explain that?
Click to expand...

You first.
 
chinglu said:
Interesting post.

If you will look at JamesR post #117 #2 in the stationary frame/unprimed frame, you will find an initial condition at x=-k/γ with a spooky at a distance desynchronization on the stationary frame clock which is not valid under SR.

I would be curious of your interpretation of #2 before I proceed.
Click to expand...

I don't understand where you think the spooky at a distance stuff comes in. Are you objecting to this event transformation?
2. $$(x',t') = (-k,0)$$
...
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$

It looks fine to me.

It just means that:
  • if all the primed clocks are einstein synchronized with each other,
  • and the primed clock at x'=0 and the unprimed clock at x=0 read t'=0 and t=0 respectively as they pass each other,
  • then when the primed clock at x'=-k passes the unprimed clock at x=$$-\gamma k$$, the primed clock shows t'=0 and the unprimed clock shows $$t=-\frac{\gamma k v}{c^2}$$.

Why do you think that is that valid under SR?
What do you think that SR says those clocks read as they pass each other, given the specified synchronization?
 
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Pete said:
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$
Click to expand...

To both posts above.

We have taken the unprimed frame as stationary. The time component in (2) above implies the clocks are not Einstein synched in the unprimed frame.
 
chinglu said:
The time component in (2) above implies the clocks are not Einstein synched in the unprimed frame.
Click to expand...

We've already been through all this.

Two spatially-separated events that are simultaneous in one frame CANNOT be simultaneous in any other frame. Do you agree? Yes or no?

Therefore, if two events occur simultaneously in the primed frame (as your problem specified), then they cannot occur simultaneously in the unprimed frame. Do you agree? Yes or no?
 
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James R said:
We've already been through all this.

Two events that are simultaneous in one frame CANNOT be simultaneous in any other frame.
Click to expand...
More specifically (I think):
  • If two clocks, T1 and T2, are:
    • Einstein synchronized with each other, and
    • T1 is present at an event S1, and
    • T1 is present at an event S2, and
    • the time T1 records for S1 is the same time that T2 records for S2,
  • and if two more clocks, T1' and T2', are:
    • Einstein synchronized with each other, and
    • T1' is present at S1, and
    • T2' is present at S2,
  • Then the time T1' records for S1 must be a different time that T2' records for S2.

This is clearly much more wordy, but it avoids misunderstandings about what "simultaneous" means.
 
I can't understand why this is so complex.

We are taking the view of the unprimed frame for this exercise.

When the origins are the same, t'=t=0 at the origins.

t=0 in all parts of the unprimed frame.

You two are confused.
 
chinglu said:
I can't understand why this is so complex.

We are taking the view of the unprimed frame for this exercise.

When the origins are the same, t'=t=0 at the origins.

t=0 in all parts of the unprimed frame.

You two are confused.
Click to expand...

well, (x', t')=(-k,0) doesn't happen at t=0.
 
Sorry chinglu, I don't know what you're thinking.
Perhaps you could explain your difficulty with James's post #117 more explicitly?
 
Let's get factual.

Any moving clock coming toward the rest origin is time expanded, not time dilated.

Any moving clock moving away from the rest origin is time dilated
 
No, you haven't explained it. James's post is clear, your objection is not.

To spell out the step you are objecting to:

  • if all the primed clocks are einstein synchronized with each other,
  • and the primed clock at x'=0 and the unprimed clock at x=0 read t'=0 and t=0 respectively as they pass each other,
  • then when the primed clock at x'=-k passes the unprimed clock at x=$$-\gamma k$$, the primed clock shows t'=0 and the unprimed clock shows $$t=-\frac{\gamma k v}{c^2}$$.

Where is the 'spooky action at a distance'?
 
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