Questions about Special Relativity

[5614]

Additionally' date=' let them be in deep space, free of all external forces.

 

I claim that if the two clocks are synchronous at one moment in time, they must remain synchronous.[/quote']

But Swansont said that they were moving relative to one another, so sure they may not be accelerating (free of all external forces), but time dilation applies to clocks moving at different speeds.

[Johnny5]

A and B are in inertial frames so they are not accelerating' date=' however they are moving relative to one another, so e.g A is moving at 10mph and B at 20mph relative to me... so from A's frame B is moving at 10mph and from B's frame A is moving at -10mph.

[/quote']

 

Are you in an inertial frame?

[5614]

Let's just say I am for simplicity's sake! (plus it's an example to demonstrate something - if you really don't like it then change "me" to "an inertial frame" but I think you were just asking.)

 

Otherwise it's like, I was driving at 1000mph past a speed camera which takes at 30mph and it didn't flash me (and the camera works) why? Because I was moving at 1000mph relative to a passing commet in outter space travelling in the opposite direction to me.

[Johnny5]

Let's just say I am for simplicity's sake! (plus it's an example to demonstrate something - if you really don't like it then change "me" to "an inertial frame" but I think you were just asking.)

 

 

I am willing to follow this out with you, so that I can see what it is you want to demonstrate. But explain things more. Apparently there are three frames?

[5614]

Apparently there are three frames?

That confused me a bit. What do you mean?

 

There's A, B and me or inertial frame.

 

A + B was introduce by Swansont to demonstrate time dilation, I wanted to understand why you where questioning swansont so went through his post, to make it clearer I added speeds in (just as an example - post #25), just to show how A and B can be inertial and moving relative to each other I just said that they were both moving relative to an inertial observer (which I referred to as [1]"me") and relative to each other, frames [2]A and [3]B... what dont you get?

[Johnny5]

That confused me a bit. What do you mean?

 

There's A' date=' B and me or inertial frame.

 

A + B was introduce by Swansont to demonstrate time dilation, I wanted to understand why you where questioning swansont so went through his post, to make it clearer I added speeds in (just as an example - post #25), just to show how A and B can be inertial and moving relative to each other I just said that they were both moving relative to an inertial observer (which I referred to as [1']"me") and relative to each other, frames [2]A and [3]B... what dont you get?

 

I think I get it all, but just to be sure.

 

Let us define frame 1, to have its origin permanently located at the center of mass of Clock A. Let frame 1 currently be inertial, which means that Newton's laws are currently true statements in frame 1.

 

Let us define frame 2, to have its origin permanently located at the center of inertia of Clock 2. Let frame 2 currently be inertial, which means that Newtons laws are currently true statements in frame 2.

 

Now, let there be a third frame (which you are at rest in), which is stipulated to forever be an inertial frame.

 

Now, let both clocks be moving through the coordinates of the third frame.

 

If they have the same velocity vectors in the third frame, then they are at rest relative to each other.

 

That's not what swansont wants. The two clocks need to be in relative motion, with the relative speed being denoted by v.

 

Ok so...

 

A and B are in inertial frames so they are not accelerating, however they are moving relative to one another, so e.g A is moving at 10mph and B at 20mph relative to me... so from A's frame B is moving at 10mph and from B's frame A is moving at -10mph.

 

Let clocks A,B currently be free of external forces. Therefore, if they are at rest in the third frame, then they will remain at rest in the third frame. If they are moving in the third frame, then they will continue to move in a straight line at a constant speed in the third frame, until such moment in time as they are acted upon by an external force.

 

Can you be more precise as to the direction the two objects are moving? Just so that I can have a clear picture of the trajectory of both clocks in the third frame please.

 

Thank you

[swansont]

This is impossible.

 

Not only is it not impossible, the symmetry of the problem requires it, as long as there is no preferred frame of reference. And that particular argument was answered >100 years ago.

[5614]

Now' date=' let both clocks be moving through the coordinates of the third frame.

 

If they have the same velocity vectors in the third frame, then they are at rest relative to each other.

 

That's not what swansont wants. The two clocks need to be in relative motion, with the relative speed being denoted by v.[/quote']

If the clocks move through the 3rd frame it could get confusing.

Since when were they at rest relative to each other?

Agreed, they need to be in relative motion, ok, lets call that relative speed +/- v

 

Direction is irrelevant, but if you want to visualise imagine me standing in space (I am inertial, forever at rest).

 

I am by the side of a straight 100m race track in the spectator stand and these two clocks A and B are racing past me.

 

B is winning the race, it is moving at 20mph, where as A is only moving at 10mph (fat lump of a clock can't run very fast!!!!)

 

These clocks are moving relative to each other (at either +/- 10mph)

If A and B are in inertial frames and moving relative to one another[/b'], each will measure the other's clock as moving slow.

Because of this speed difference between the two clocks time dilation applies, their clocks move at a different speed

 

Now Swansont said "each will measure the other's clock as moving slow" but surely B (moving quicker) clock would be going slower relative to A, consequently A's clock would be moving faster relative to B's clock[/u'].

To give you which clock is moving faster I had to give each clock a speed, which is where the 3rd frame came in.

[swansont]

A and B are in inertial frames so they are not accelerating' date=' however they are moving relative to one another, so e.g A is moving at 10mph and B at 20mph relative to me... so from A's frame B is moving at 10mph and from B's frame A is moving at -10mph.

 

Because of the speed difference and SR's time dilation A's clock and B's clock will run at different speeds.

 

Now Swansont said [i']"each will measure the other's clock as moving slow"[/i] but surely B (moving quicker) clock would be going slower relative to A, consequently A's clock would be moving faster relative to B's clock.

 

Not in A or B's frame - each will see the other clock run slow. A and B are not moving at different speeds, according to each other. For either observer, they are at rest and the other person is moving at v.

[Johnny5]

NIce and slow.

 

You are in the spectator stand with a stopwatch, and there is a race between fat clock A, and skinny clock B.

 

Skinny clock B is going to win the race, because he can run faster, and you who are in the stands can see that clearly.

 

The race is the 100 meter dash, and there is a tape measure, with marks on it, and the tape measure is at rest relative to you, and you are just sitting in the stand, with your stopwatch.

 

Now, skinny clock and fat clock start the race neck and neck, at the point zero on the tape measure.

 

But thereafter, skinny clock is ahead.

 

At the moment the race begins, skinny clock reads zero, and fat clock also reads zero.

 

Then you fire the starting gun, and they both take off simultaneously in your frame of reference.

 

OK SO... lol

 

Let the tape measure be along the x axis of your frame.

 

The rest frame of each clock, was stipulated by swanson to be inertial. That means that in either of their frames of reference, you are moving in a straight line at a constant speed, and vice versa.

 

So each of these clocks is moving in a straight line at a constant speed in your frame, and you are moving in a straight line at a constant speed in either of their frames.

 

Skinny clock can run down the track at a speed of 20 meters per second.

Fat clock can only run down the racetrack at a speed of ten meters per second.

 

Here are their velocities using vector notation, as defined in your frame:

 

V_s = 20 i^

V_f = 10 i^

 

Now, their relative speed is 10 meters per second, so the clocks are moving apart in your frame, at a rate of 10 meters per second.

 

Now, let skinny clocks rest frame be such that his x axis is parallel to your x axis, and let fat clocks rest frame be such that his x axis is also parallel to your x axis, and therefore parallel to skinny's x axis as well.

 

Here is what fat clock sees:

 

Fat clock sees skinny moving with velocity 10i^ in his frame.

 

Here is what skinny clock sees in skinny's rest frame:

 

Skinny sees fat clock moving away from him with velocity -10i^.

 

So all three of you agree, that the relative speed between the two clocks is 10 meters per second.

 

The way I've set things up, you can formulate true statements in any of the three frames, that you wish.

[5614]

For either observer, they are at rest and the other person is moving at v.

Woa, you just set off a whole argument in my head!

 

How can each frame know that they are at rest?

 

It's a fairly big assumption, or is it a standard assumption that in physics if there are 2 frames and I am one of them I can say (when referring to time dilation) I am at rest and the other is moving at v?

yeah, i spose that makes sense!

 

each will see the other clock run slow

I've redrafted this post like 10 times now... you can't add a third frame to time dilation because it really gets confusing and harder to explain that quote... but I do think I get it now.

[5614]

So all three of you agree, that the relative speed between the two clocks is 10 meters per second.

OK, firstly ignore the 3rd frame because it just gets confusing.

 

If you were frame B you would see A moving away from you... you assume you are at rest and A is moving.

 

If you were frame A you would see B moving away from you... you assume you are at rest and B is moving.

 

Consequently each will see the other clock run slower... because A thinks B is moving quicker and B thinks A is moving quicker.

[Johnny5]

Consequently each will see the other clock run slower

 

 

 

Prove this.

 

I expect to see the time dilation formula utilized.

 

 

PS: I also know that the third frame is unnecessary because all three frames are inertial, so there's a bit of redundancy (which ok but just not necessary), so just use the two frames, the rest frame of skinny clock, and the rest frame of fat clock, which were stipulated to both be inertial reference frames.

 

Let v denote their relative speed.

 

(which happens to be 10 meters per second, everyone is using the same rest units for distance and time measurments have to say that somewhere)

[5614]

well the time dilation formula is:

 

used an image I found just because latex isn't working atm and I'm not that good enough to get delta and square route!

 

So we can all put in the numbers!

 

V being the variant, we can see that the output will be different for each clock.

 

And no I cannot derive that [time dilation] formula!!!!!

look here:

http://www.physicsforums.com/archive/topic/t-29984_Trying_to_derive_the_time_dilation_formula.html

for that

[Johnny5]

Unfortunately the image didn't come up, so now what?

[5614]

It does for me!

 

Go to:

http://members.tripod.com/wmhxbigguy/pics/formula_t.gif

 

It should work!

 

or:

http://home.usit.net/~cmdaven/timedil1.gif

http://www.stardestroyer.net/Empire/Science/Science2.gif

http://www.krysstal.com/images/formulas_time.gif

[Johnny5]

Yes the time dilation formula.

 

Dt = Dt`(1-vv/cc)^-1/2

 

Where v is the relative speed between fat clock and skinny clock, which is 10 meters per second if i do recall, and c = 299792458 meters per second.

 

That's the formula right?

[lxxvii24]

the lenght contraction phenomena is only the consequence of travelling at relativistic speed.

[5614]

the lenght contraction phenomena is only the consequence of travelling at relativistic speed[/b'].

Well you can't have absolute speed... so what other speeds are there?

 

Dt = Dt`(1-vv/cc)-1/2

 

Where v is the relative speed between fat clock and skinny clock' date=' which is 10 meters per second if i do recall, and c = 299792458 meters per second.

 

That's the formula right?[/quote']

Yep, that's the one, although the ^-1/2 only applies to the part in the brakets, not the delta t`

[lxxvii24]

common sense is indeed common, ity is only the application of it that is not common

[Johnny5]

the lenght contraction phenomena is only the consequence of travelling at relativistic speed.

 

This is totally incorrect. In the time dilation formula, the symbol v is the relative speed between the centers of inertia of two things.

 

So...

 

V could equal zero, in which case the two things aren't moving relative to each other, we would say they are at rest relative to one another.

 

Or, v could be 10 meters per second, as in the case we have here, where two clocks (skinny and fat clock) are running the hundred meter dash.

 

You can answer your own question using the time dilation formula. v is the relative speed. There is no mathematical reason you cannot put v=0 in that formula.

 

One problem would be in the case of division by zero error.

 

Mathematically, that would occur if the relative speed v, between the centers of inertia of the two clocks was equal to c. In the case where v=c you have division by zero error UNLESS... the numerator is simultaneously zero.

 

According to the formula, if the relative speed v exceeds the speed of light, then "amount of time" becomes imaginary, which is nonsensical.

[lxxvii24]

my friend common sense is common, it is only the application of it that isn't common

[5614]

v^2 will never be bigger than c^2 therefore vv/cc will never be bigger than 1 and so 1-vv/cc will always be between (and excluding) 1 and 0.

[Johnny5]

v^2 will never be bigger than c^2 therefore vv/cc will never be bigger than 1 and so 1-vv/cc will always be between (and excluding) 1 and 0.

 

Why can't v be greater than c?

[5614]

Because a clock must have mass therefore cannot exceed the speed of light.

 

As we're dealing with SR which is part of physics lets just agree with the laws of physics, ie. nothing with mass can go FTL or c.