spunnery

A running track to be built in advance to be directly in the path that the Earth spins so that if a race was held on the track (doesnt need to be an oval) that the runner could in fact run a faster time if they ran in the opposite direction of the Earths spin.

 

Or would factors such as gravity and other forces relative to our atmosphere make this impossible and wouldnt give the possibility of running faster.

In any case ,the runner is having a momentum(even if he stands still)equal to that of the spin of earth.in other words,he is moving back with a velocity equal to the spin of earth.So no runner can run faster ,if he ran in opposite direction of earths spin.

if you watch the runners from space,you will see the runners running forward,but they are moving backwards,together with earth,Only their relative position with the ground below have been changed.

A, B and C are in the same position, relative to each other. There is no A', B' and C' in that frame. If you have an absolute frame, you would expect the proposed change, but this is not observed experimentally, so it is falsified.

 

Is there point to any of this?

Sketch2 and sketch 3 are describing two different cases.sketch 2 is the case ,where there is no movement of ship.Sketch 3 is the case if there is movement.

if observed time (from B) ,for both the clocks at A and C are same,the result shows that there is no motion of ship.(this is considering that light is travelling at constant speed)

On the other hand,if light is travelling at a relative speed with source,even if ship is moving,the clocks will show same time.

so either one can be true.

Waaaay back in post #6

 

 

 

But your diagram and post imply that the speed of the clocks with respect to some other frame need to be taken into account. And that's wrong.

Sorry,I am discussing now about post#12.I made a bad notation A',B',C',which is confusing you.A',B'and C' are the position of A,B and C respectively ,before 't' seconds.please see both are same frame only-sorry again

No, it isn't. SR iteself is a self-cosistent result of assuming c is constant. It's derived from that and some mathematical definitions. Meaning it will hold if that postulate is true (and if the laws of physics don't change). All you can show is that the assumptions don't model the real world, and that requires a physical experiment. A gedanken experiment will include what you think nature will do. So if there is a disagreement, all that means is that you predict nature will disagree with theory, but it doesn't disprove anything until you do the experiment to confirm it.

 

So if you are describing a situation that has already been investigated, like moving clocks, and you have a problem, it's because you have misapplied the theory. If there was a delay as you describe due to motion through an absolute frame, synchronizing clocks at noon vs midnight would be different because of the change in velocity due to rotation of the earth. This is not observed to happen.

 

 

This isn't an issue of prejudice. It's about burden of proof and exisitence (and quality) of evidence.

Please note that i mensioned the time delay,as observed from point B.but actual time at all the three points remain same(there is no time delay).

And for the mathematical part,it is clear that i assume the first postulate of SR-light appear to be travel at constant speed for any observer-(i.e,velocity of light is not affected by the velocity of source).So first agree with sketches

(if don't tell me why?)

remaining calculations are simple mathematics only.please go through it again and tell me preciselly in which step ,i have made a mistake.

As per my argument,no time delay due to change in velocity,so there is no point in arguing about the time delay due to rotational velocity.

 

A delay willbe observed,which is equivalent to the time taken for the light reflected fromm the needle of clock ,to reach a far away observer-Analogy -a far away star we see now is a star of past.

Appologise for harsh meanings,if any for my sentences-i am not good enough in english language.

If the clocks are all moving with respect to some other observer, so that they are not moving with respect to each other, they will all agree with each other in their own frame.

 

The basic flaw is this: special relativity is a self-consistent theory. There is no thought experiment you can do to show that it's wrong, unless you misapply relativity. If you come up with an inconsistent answer, then you have made a mistake. The only way to falsify it is through actual experiment.

 

IOW, gedanken experiments show you how SR works. They won't show you that it doesn't work.

My first request is ,not to see the problem with a prejudice.

secondly,special relativity and general relativity itself are result of thought experiments,which have to be proved yet (atleast a non negligible percentage beleive so).

Again ,the word gedanken,literally means as 'thought',in german language.Which you are telling ,will show me,how SR works.So a gedanken experiment is quite enough to falsify it also.

 

finally,the experiment i mentioned can be done practically on a small space ship,by reflecting,the lights to travel a considerable distance, using mirrors fixed at an angle to the side walls of ship.

The mirrors are moving, though. If the separation distance is d (which is length-contracted from the rest length) and the clock is moving at v, the forward ray will move a distance ct = d + vt in time t. (the orginal d and the extra distance the mirror moves, which is vt) The return will be ct' = d - vt'

 

 

That gives you (assuming my math is correct) a distance of [math]\frac{2d}{(1-\frac{v^2}{c^2})}[/math] (and d is length contracted)

Sorry for delay.and sorry for a useless attachment.

Here are some interesting stuff on the subject.(please refer sketches)

Consider you are in a space ship.Say it is 300,000km long(For the experiment ,it is not mandatory to be the length of ship to be that much.You can deflect the path of light using mirrors to achieve desired length).

Refer the sketches in attachment.

Synchronise three clocks from position A

take one clock to position B midway and another to position C (front end)motion is considered to be in AC direction.

Now see the sketch 2,if there is no motion,as observed from B,you will see the clocks at A and C showing a time of T-0.5 seconds with respect to B's clock.(150,000/300,000=0.5).

Now see sketch 3. it is self explanatory .is it interesting?

if the light is travelling at constant speed,this will be the result.

if so ,this is an experiment which can be done from an inertial frame ,to show that frame is moving at a constant velocity,u.

Otherwise,special relativity is flawed.

Correct me,if i am wrong.?

LIGHT.pdf

The mirrors are moving, though. If the separation distance is d (which is length-contracted from the rest length) and the clock is moving at v, the forward ray will move a distance ct = d + vt in time t. (the orginal d and the extra distance the mirror moves, which is vt) The return will be ct' = d - vt'

 

 

That gives you (assuming my math is correct) a distance of [math]\frac{2d}{(1-\frac{v^2}{c^2})}[/math] (and d is length contracted)

xxx

time in action.pdf

The orientation of the clock does not matter. If you get a different answer for the longitudinal clock, then you did the derivation wrong.

 

 

 

Observation or measurement is what matters. If you measure it to be shorter, it is shorter, in your frame of reference.

if the length is really shortened,that is the answer for cancellation of time dilation,in a longitudinal clock.no need to derivate the whole formula.

otherwise,i will just derive it here,correct me if i am wrong?

 

the total horizontal distance travelled by clock = vt

distance travelled by reflector when the light ray from sorce hit mirror = vt/2.

total length of forward path of ray = h(distance between source and mirror)+(vt/2).

 

total distance travelled by ray in backward path will be = h -(vt/2)

 

 

Hence total distance travelled by ray will be '2h'.which is same for a clock at rest.But, h'(total path) for perpendicular moving clock is 2 multitplied by sqrt of [(vt)^2 + h^2]

So it is clear that ,no time dilation in the case of longitudinal clock.this is the other side of the reason why length contraction happening only in the direction of motion.

You take into account the path delay from finite speed of light when you sychronize clocks. You adjust the reading by t=d/c.

 

Time dilation is different — the rate of the clock is changed, not just the reading. Always lagging by a second is not running slow. Running slow means continually losing time over repeated intervals. My watch, for example, runs slow as compared to the US Master clock — it loses about 1 second per day. So if I synchronize it today (taking into account path delay), it will be behind by a second tomorrow, and two seconds the next day, and three the day after that.

 

Of course, if you can measure the change in rate, the clock becomes useful again, which is what happens with GPS satellite clocks. They run fast (from kinematic and gravitational dilation effects), but the synthesizer frequency is adjusted to compensate.

In time dilation,the rate of clock is changed -because the path of light(and distance increasaed) is changed.This is very clear from the derivation of time dilation using pythogorus theorom(I know the lorentz transform derivation is more complicated).In the above derivation if you keep the clock horizontal(ie, instead of prependicular to the motion,you keep clock parallel to the direction of V),you can see the effect of V is nullified(please imagine the total path of light).means clock is running at a same rate as if it was at a rest frame.Does it means,for time dilation ,to occur you have to keep the clock perpendicular to motion

 

Same in the case of length contraction also,in a moving frame,if you use a ruler to measure length,the ruler also is subjected to the shorteneing (in appearance)and finally you will measure the same distance.But if you try to measure a distance in a moving frame,from a frame at rest,of course it will appear to you that distance is shortened.this is only observation and the true thing is distance remains same.No shortening

This would apply to any transverse motion if it were correct. I mean, "vertical" really has no meaning on a spaceship.

 

So it sounds like you have to anticipate where the target will be in order to properly align the trajectory. You've introduced a preferred frame, and an actual experiment testing this would fail. e.g. lasers traveling a significant free path (several meters) and then aligned into micron-sized optical fibers should become misaligned after 12 hours because the alignment relative to the motion of the earth around the sun has changed.

By vertically ,i mean perpendicular to direction of motion.

Could you please explain me regarding the laser thing(i appologise ,i am only playing with some basics)

Anyhow i will try to explain my point again.

Two synchronised clocks are kept 300,000 km apart in a straight line at A and B.Observers at A & B will see each others clocks are one second behind their time.[time taken by the light (velocity is approximate only)to travel from A to B and vice versa].Now consider A is travelling with his clock towards B upto C (150,000 km.from A)No matter what speed A is going,When he reach C he will find now B's clock is only half second behind him.For A ,B's clock appears running faster while he is travelling and vice versa.

A's speed will only change the rate of change in B's clock from 1 second lag to half second lag.So basically the result depends on the position of A at the instance.

Consider that both are travelling towards each other and with high velocity.things become more complex . Here relative position at the instance is more important to get in to a conclusion about the results on observervations made by both A and B .But you will see that at any instance,the time elapsed by both clocks are same.So my argument is that use Lorentz transform to obtain the results.but the spacetime curvature and time dilations are only the observations but not the results.

surely as the ball is horizontally at rest in his reference frame he would see no relativistic effect?

of course the ball is horizontally at rest in his reference frame(this is actual position),but he is not seeing it .he will see the ball at a position where ball was before 't' seconds ,depending on his speed.

Is lorentz transforms are the complex version of the situation ?Note that a correction is required to find the exact position ,from the observed position.

some imaginations.

Me and my friends were travelling in a rocket with a constant velocity towards a planet.nobody except me knows that rocket was travelling towards planet.There was no background stars or anything in the sky visible through the front glass,to make us aware of the motion.

My friend A told me "see it is a special kind of planet .it is expanding"

B told me "A is wrong,That planet is moving towards us".

C told me "no no we are moving towards the planet".

How can you say A is wrong?.Only because i know the rocket is moving towards the planet .But can i say A is completely wrong ? No.Even we are moving towards and if planet is expanding as well,we will see the same scenario.To say A is absolutely wrong,I need more datas like distance ,diameter of planet,velocity of rocket etc.

If we know the rocket is at rest,then either A or B or both of them are correct.

What i want to say is that in such problems, before coming to any conclusion about what you have observed,you have to put in consideration with all other datas , the relative position and movement of the observer.

 

Relative position of the observer ,can cause a major difference in observations made.So you have to make proper corrections to the equations ,instead of thinking wild about a warped spacetime.

 

A simple example,

A man throws up a ball vertically from a ship moving at a constant velocity.if he is moving at a small velocity compared to that of velocity of light.For him ,the position of ball is always vertically above his head.This is because the horizontal velocity of ball and ships velocity is same.Finally fo him, the ball will reach back to his hand after executing a vertical path.

 

What happens,if the ship is moving at a velocity,nearer to that of light.this time also,the balls position will be always vertically above his head as before and because of same reason.Butin this case, before the light reflected from ball,reach him,he will move ahead horizontally to a new position ,where the light pulse will reach him. So he will see the ball at an angle behind him.Finally he willl observe that the the ball is taking an inclined path backwards and coming back to his hand through an inclined path.

From this,if the observer deduce that if he throw a ball vertically up,it will move up in an inclined path,it will be total blunder.So what he can do is to give proper corrections to the equation to make the path straight.And this is in true the position of ball.

 

I explained all this because,i want to say that what Lorentz transformation is doing ,is only the necessary corrections to the observations,considering the relative movement and position,of the observer.

There is no such time dilation or length contraction,but only necessary correction according to the relative position and movement of observer taken in to consideration.

Anything more interesting?

stop you are making a big mistake.

the cosmological redshift is by definition NOT a Doppler effect and is calculated differently from the relativistic doppler formula

 

 

in your initial post you asked about the DOPPLER shift, so that has nothing to do with cosmology and the expansion of space

 

so I agree with everything swansont said. that is about doppler shift and has to do with motion in our reference frame, galaxy approaching or going away (and incidentally can also have sideways motion but this does not contribute)

 

===============

the cosmological redshift is a completely different ballgame

there is a lot of stuff out there which we actually observe which is receding from us at 2 or 3 times the speed of light because of the expansion of distances

 

there is no doppler formula that can translate 2c into a redshift:-)

 

you should try one of the cosmology calculators. google Ned Wright and check out his

 

if you find Wright's calculator and play aroundwith it and decide you like it, let us know and I will dig up a link to Siobhan Morgan's

 

these calculators do things like convert a redshift into a distance and a time back in the past and a recession speed. you can play around with them and see what is happening at various redshifts and get a feel for the universe.

Thank you .(and incidentally can also have sideways motion but this does not contribute).This was the answer i was looking for.Sorry for the ignorance in the subject.

If the speeds are not relativistic then there will be no shift. If v is large enough, you will see a transverse Doppler shift, from the time dilation of the moving object.

Answer not seems to be specific to my question.(may be my question is not clear).The acceleration of expansion of universe is calculated from the doppler shift of spectrum of radio waves emitted from the galaxy.Do we give any consideration (or say correction)for the direction of motion of galaxy with respect to earth(or in general say with respect to our galaxy).And if not doing so,does it make huge errors.

Could anybody who is familiiar help me with some details of doppler red shift?

In the calculation for velocity of an object(say galaxy) ,do we consider the direction of motion of body.I will explain

Object receding will form a red shift.

Object approaching will form a blue shift.

What about an object moving at an angle with the observer.Of course the value of shifting will be different.

Do we give any consideration for this while calculating velocity?

for example,

If the observer is at right angle to the direction of object,he will recieve signals without a shift????

You haven't changed any physical laws.

 

Light speed is constant to any inertial observer. An accelerating observer will not, in general, measure the speed of light to be c.

So why the paradox,Inspector chasing the light at .99c observe the light travelling at c?.This can be happen only if source of light is also moving at 0.99c,Otherwise he will observe the light with just 0.01c (same as the difference between the ball dropped from bridge and ball dropped from inside the train).

Correct me ,if i am wrong?

Otherwise in general ,we can say all the physical constants remain same to any inertial observer

That sounds right. What's the issue with this?

Issue is that physical laws are not changed in any frame.How comes light can break the law(to be constant to all observers)?.Why the simple logic in this problem not applicable to light?

Does any physical laws are changing when you travel at a speed nearer to light?.I think only thing is you have to rewrite the formulas according to the position (space co-ordinate at the instance)of the observer.

It's not the same experiment if the train is moving relative to the bridge in one case, and not moving relative to the bridge in the other.

Ok i agree.Shall i try to say with one example,that accelerating frames are also inertial frames.But only thing is that we have to make corrections to the results,considering the aceeleration.

for example;

Case1 - train is moving at a constant speed.

A person inside the train dropped a ball verticaly down.At the instance he dropped the ball,both ball and train have same horizontal velocity.Gravity pull the ball down.Resultant of the horizontal velocity and vertical gravity will bring the ball through the diognal of paralellogram,.In the mean time floor of train (with the person)will travel equal horizontal distance,as ball has travelled.So the ball will hit the floor of train verticaly down the person,since the final position of the person is changed.

Case-2 -the train is moving at an acceleration.

The ball dropped,in this case will have the velocity ,V1 of the train at that instance(inertia?).Now the ball will travel through the diognal of parralellogram drawn with v1 and gravity.but the train ,is acellerating(it will have a greater velocity and will be displaced horizontaly more,with respect to the ball).The floor and person will move little further than the point (depending on rate of acceleration)where ball will hit the floor.So the observer will see the ball hit the floor behind him(and this is true).Does any physical laws changed here?.only thing is you have to give appropriate value to the final position of the person(observer).

consider a train moving at a constant velocity. let the top of the train be open.drop a ball verticaly to the train from a bridge ,while train is crossing the bridge.What will be the path ball.Again drop the ball when the train is stopped below bridge. In both cases the ball will hit the floor of train at different places.Can it be considered as a physical experiment showing different results of same experiment for a body at rest and a body moving at a constant velocity?

Now can we call the moving train as an inertial frame.

We're not supposed to give homework answers. We can help them do it themselves, but we are not to do their homework for them. You can give equations and hints, but not answers.

Sorry I was not aware about that.really sorry

Hi,

 

I wonder if someone could help me with two motion problems.

 

The first one is:

 

A car can accelerate at 1.9m.s-2. It is initially travelling at 19m.s-1 when it passes through an intersection and it then accelerates for the next 85m. What is the final velocity of the car?

 

And the second one is:

 

A car is travelling at 95km/h for 130km, and then slows down to 65km/h. It takes 3 hrs and 20 mins. How far has the car travelled?

 

If someone can help, I would be very grateful!!

 

God bless.

The first answer is 31.709 m /sec.

Second answer is 257.719 km

I can't seem to find one decent animation of the earth / moon orbit, that will show spunnery exactly what's being said here.

Did you say i am right or wrong? I couldn't get what you mean.

gravity, however the person standing on the platform wil see the ball on the train falling slower.

 

.

Gravity of course!.The person standing on the platform will see th ball on the train falling slower? Never .He will see a different path for the ball ,instead

No, the twins won't be the same age, The twin that underwent an acceleration changed from one inertial frame to the other. That breaks the symmetry. Since acceleration is not relative you can tell, between the two, whose clock is correct.

 

This is explained in some detail at many places on the 'net.

Sorry i was late on the subject.Here i am twisting little and i would like to ask you to explain some mathematics about the travelling train.

 

The same old experiment(tired?).Consider a train moving with a uniform velocity 'V' and you are standing inside the train.Now you are dropping a ball of mass 'm' from a height 'h'.You will see the ball dropping vertically down to the floor of train .

The same result will happen if you do it on the platform also.

I would like to hear from you ,why this is happening so?I am looking for some mathematical explanation rather than the terms Inertia ,parabolic path etc,etc.

You don't need a personal ID. Click on a member's name (Bignose invited you to send him a message) at the top left of any of their posts and you will get a menu allowing you to send a Private Message. Type it out like an email and you can send Bignose a message that won't be seen by anyone else.

Thank you