This is in parallel with my post on Tensors. I would like to use this thread as a place to discuss and learn General Relativity. I'm not looking for debates, because you've first got to know at least the basics before you can argue.

I'll pose a question here. Consider a frame of reference which has a uniform acceleration 'a' (w.r.t. some inertial frame). From what I've read, observers at different heights in this frame will have different rates of time in their clocks. It's not because of anything like variation of gravitiational field or anything, since we're only considering a uniformly accelerating frame of reference. I don't know if this makes much sense physically. Could someone give a good argument why there should be this time dilation? And could you also please show that math?

This is where I learnt of this effect, but it doesn't give the mathematics. And, when I said a physical reason I meant one which was different from that given in this site.

<A href="http://www.upscale.utoronto.ca/GeneralInterest/Harrison/GenRel/TimeDilation.html">Gravitational Time Dilation</A>

This is the argument given by Einstein in 1911. It is, by his own explicit admission, approximate, but it seems to be what you are asking. A qualitatively similar conclusion can be derived directly from the field equations.

Consider a system of coordinates K₀ defined by rigid, material axes X,Y,Z. Next consider a second such system K, similarly defined. Let there be in K, rigidly connected to the Z axis, a source of radiation S₁ at z₁ and a detector S₂ at z₂ at a distance h from S₁. At time t = 0 let S₁ emit a finite quantity of radiation energy E toward S₂. Also at t = 0 let K move relative to K₀ in the positive Z direction with constant acceleration &gamma; such that, over the minutely small time interval under consideration v = &gamma;t. We expect the radiation to arrive at S₂after the time interval h/c. But by this time S₂is moving relative to K₀with velocity v = &gamma;(h/c). So by the special theory the amount of radiation arriving at S₂, E’ , is greater than E by approximately v/c. So
E’ = E(1 + v/c) = (1 + &gamma;h/c²)

By the principle of the equivalence of acceleration and gravity the same relation must apply when gravity (in the negative Z direction) replaces acceleration. Writing the gravitational potential as &Phi; we have
E’ = E(1 + &Phi;/c²);

Now the same argument is used to assert that the frequency of the light, f₁, emitted from S₁ is Doppler-shifted on its arrival at S₂ as

f₁ = f₂(1 + &Phi;/c²)
This can only be explained by asserting that clocks at S₁ and S₂ run at different rates.

This is in parallel with my post on Tensors. I would like to use this thread as a place to discuss and learn General Relativity. I'm not looking for debates, because you've first got to know at least the basics before you can argue.

I'll pose a question here. Consider a frame of reference which has a uniform acceleration 'a' (w.r.t. some inertial frame). From what I've read, observers at different heights in this frame will have different rates of time in their clocks. It's not because of anything like variation of gravitiational field or anything, since we're only considering a uniformly accelerating frame of reference. I don't know if this makes much sense physically. Could someone give a good argument why there should be this time dilation? And could you also please show that math?

No, the observers at different heights in this uniformly accelerating frame will not have different rates of time. The different time rates happen only in gravitational fields. Equivalence principle is applicable only over small distances within this frame.

Are you sure? That's not what I've read. See Quarkhead's post above. Can you provide some explanation to support your argument?

Everneo: Precisely. It applies only in the infinitely small interval that v = &gamma;t, as Einstein points out. This is the point that Zanket seems not to understand.

Rosnet: Note that Einstein is using the special theory for an accelerating frame. This is only legitimate when, as repeatedly stated, that v = &gamma;t. This is why it is an approximation.
Perhaps of more interest in his 1912 paper is the relation
E' = E + E&Phi;/c². I quote:
"....we have to ascribe to the energy E, before its emission...a potential energy du to gravitation, which corresponds to the gravitational mass E/c²"

This is, of course, exactly the result he obtained from the special theory. Moreover, it forced him to conclude that not only matter but also energy contributes to the gravitational field. This was a crucial key for formulation of the general theory.

From what you've posted, I understand that time dilation occurs for clocks at two different heights in a lift accelerating upwards, right? And also, since the Equivalence principle is valid only locally, in a gravitiational field, there will be an additional factor to time dilation?

You seem to confuse between (1) the clocks in a gravitational field (say, fixed on earth's surface at different height) and (2) the clocks fixed in a long rocket moving with acceleration 'a' (say, 9.8m/s^2).

Case :

in (1) the clocks have different time rates - due to gravitational field gradient

in (2) the clocks have the same time rates - acceleration due to motion does not cause any difference in time rates of the clocks.


this way you can say whether you are moving with acceleration 'a' or you are just experiencing gravity. It is not the failure of equivalence principle but its restricted applicability over small distances rather. This is what explained in the link you provided.

Quarkhead :

You are right. I had pointed out this point to Zanket in the other forum www.thescienceforum.com few days back, ofcourse fruitlessly.

I'm saying that the clocks in (2) should also dilate. Look at the expression Quarkhead gave:

E’ = E(1 + v/c) = (1 + γh/c2)

It depends on 'h', the height. And this was my original question. Can anyone say why this should make sense? I'm not saying that it doesn't, and lots of things in relativity don't make sense to the intuition, but all I'm asking is whether the height has any special property.
<Br>
And also, isn't there anyone else here who's learning GR. Don't think about the first question alone. Post your problems too.

Again you are confusing..

Quarkhead used 'h' only in explaining doppler-shift due to motion (the clocks move over a small distance); not to say there is difference in time rates.

similar shift occurs in when the clocks don't move but in a gravitional field - hence the doppler-shift is due to the difference in their time rates.

Have you tried the link I've given in my second post? According to the arguments given in that page, there should be time dilation in a uniformly accelerating frame of reference. Forget gravity for a while.

The link talks about clocks fixed at different heights on earth's surface. Either your initial post is not clear or you misunderstood the link.

Yes, but it says a small room. It's talking about a local space, where gravitiatoinal acceleration is approximately uiniform. And also, the actual argument is based on the accelerating lift. Where does a variation of force come in there.
<Br>
Where <I>is</I> Quarkhead?

Yes, but it says a small room. It's talking about a local space, where gravitiatoinal acceleration is approximately uiniform. And also, the actual argument is based on the accelerating lift. Where does a variation of force come in there.

Where are the clocks in the room (in the link)? But they are very much there in the lift/frame in your initial post :

Consider a frame of reference which has a uniform acceleration 'a' (w.r.t. some inertial frame). From what I've read, observers at different heights in this frame will have different rates of time in their clocks.

Where in the link it is indicated that "observers at different heights in this frame (room or lift or frame or whatever) will have different rates of time in their clocks." ?


Where <I>is</I> Quarkhead?

He is neither inside the room nor inside the lift.. err..oh..sorry.. best of luck.

Perhaps if some of you were more familiar with the clock postulate, mechanical acceleration and time dilation would become more clear. There have been several
experiments to back the fact that clocks do not slow due to mechanical acceleration,
only velocity and gravitational acceleration effect clocks. I don't have time now to
look up the relevant experiments, but they were done iin both particle accelerators
and centrifuges. Here is a cut & paste and link to get you started.

This is often called the "clock postulate", but it applies to much more than just clocks, and in fact it underpins much of advanced relativity, both special and general, as well as the notion of covariance (that is, writing the equations of physics in a frame-independent way).

The clock postulate can be stated in the following way. First, we take the rate that our frame's clocks count out their time, and compare that to the rate that a moving clock counts out its time. Before the clock postulate was ever thought of, all that was known was that when the moving clock has a constant velocity v (measured relative to the speed of light c), this ratio of rates is the Lorentz factor

gamma = 1/sqrt(1-v2)

The clock postulate generalises this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, it only depends on v, and does not depend on any derivatives of v, such as acceleration. So this says that an accelerating clock will count out its time in such a way that at any one moment, its timing has slowed by a factor (gamma) that only depends on its current speed; its acceleration has no effect at all.

In other words, the accelerated clock's rate is identical to the clock rate in a "momentarily comoving inertial frame" (MCIF), which we can imagine is holding an inertial clock that for a brief moment slows to a stop alongside of the accelerated clock, so that their relative velocity is momentarily zero. At that moment they are ticking at the same rate. A moment later, the accelerated clock has a new MCIF, again one that is moving momentarily to match its speed, and there is a new inertial clock that briefly slows to a stop alongside of the accelerated clock. And again, the rates of accelerated clock and the new inertial one will momentarily be the same.
http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

Where <I>is</I> Quarkhead?He's here, dear boy, after a hard day at the coal-face. Gotta eat, shower etc. Next time I'm on, I will have had a beer or two. Let's see. I would love to have a serious chat with all you chaps about the general theory. I too am a beginner in this fied. My instinct would be to start with with manifolds. What do you, ever. and Ros. or anybody else, for that matter, think?

As long as we are talking about pre-Reimannian notions of gravity, let's just complete the picture with another of Einstiens early attempts.

Consider two sets of coordinates K₀ and K. Let their origins coincide and likewise their z-axes. Image a circle drawn around both origins in the x,y plane. Now let K alone rotate around its z-axis relative to K₀ with angular velocity &omega; = vr. Or, taking angular velocity to be be accelerated velocity, as before, &gamma; = v/t.

Now, in the infinitely small interval that v = &gamma;t use a standard measuring rod to determine in K₀ the ratio of the diameter of this circle to its circumference. It will be &pi;. Now, from K₀ determine in the same way this ratio in K. One will find, due to the familiar length contraction of the measuring rod due to its motion in the direction of travel as established by the special theory, that this ratio is greater than &pi;. Euclidian geometry has thus broken down.

Similarly, the trajectory of a particle, say in the negative x direction which, in the x,y plane of K is a straight line will appear from K₀ to be curvilinear in the x’,y’ plane of K. This curvilinear trajectory can be considered under the equivalence principle to be motion under the influence of a gravitational field. And if the "particle" is a massless instantaneous pulse of light, we must say that this is due to the curvature of space.

Or to put it another way, as light is not subject to a Newtonian gravitational "force", and always travels in straight lines, what appears to be a straight line in either system viewed from "within", will not appear to be straight in K when this line is viewed from K₀.

Right. It doesn't say that about the room. But look at what it does say.

When we pass by Clock 1, it is moving with respect to us. Therefore, Special Relativity tells us that it is running slowly relative to our clock. Similarly when we pass by Clock 2, since it is moving with respect to us it will be running slowly compared to our clock. But, since we are in free fall our speed with respect to the Earth and the two Earth-bound clocks is increasing: we are accelerating down at 9.8 m/s2 relative to the Earth. So when we pass Clock 2, its speed with respect to us is greater than the speed of Clock 1 when we passed by it.


It does say afterwards that the clock in the stronger gravitiational field will be running slower than the one in the weaker field. But wait! Suppose that the field <I>is</I> uniform. Even then, this argument should hold. There doesn't have to be variation in acceleration in order for there to be a variation in velocity. And this argument is based on the fact that the <I>velocity</I> of the (accelerating) observer would have increased by the time he reaches the lower clock. Thus, this clock should be dilated more than the first one. Or, I'll give the same example, but taking place in the lift.

The lift is accelerating upwards. (There's no gravity). Someone jumps into the lift from the top of the building. Actually, he'll just float, since there's no gravity. The lift is far below as yet. There's a clock on the top of the lift and on at the middle. He will see the upper clock pass by him at a certain velocity, and hence, being time-dilated. And he'll see the lower clock pass by him at certain velocity which is greater than the velocity of the first clock, and hence he'll see that the lower clock is slower than the firt upper one. What's wrong with this picture?

Yeah, we'll have to get into that sooner or later. But before that, someone will have to contribute to my Tensors thread. And don't forget our discussion about complex space-time. You seem to have realized the reason for our disagreement. I'm anxious to hear about that. Although I didn't think that it would come to this while making the original post. Essentially, it just shows another way of doing the math, and isn't very interesting from tht point of view. But its physical implications are intriguing.

I'll give the same example, but taking place in the lift.

The lift is accelerating upwards. (There's no gravity). Someone jumps into the lift from the top of the building. Actually, he'll just float, since there's no gravity. The lift is far below as yet. There's a clock on the top of the lift and on at the middle. He will see the upper clock pass by him at a certain velocity, and hence, being time-dilated. And he'll see the lower clock pass by him at certain velocity which is greater than the velocity of the first clock, and hence he'll see that the lower clock is slower than the firt upper one. What's wrong with this picture?

If you don't mind, let me ask a question.


Are you assuming here too uniform gravitational field for both the clocks in the lift?
If yes, assume a person inside the lift, would he/she see these clocks differ?

Yeah. I've said there's no gravity. I'm considering only an accelerating lift, and an observer who seems to be in free fall for anyone loking from inside the lift.

For anyone inside the lift the observer is freefalling, right. But my 2nd question is, for anyone inside the lift would the clocks show different time?

The freefalling observer would see the clocks as being dilated at different rates. But I see your point now. I don't know whether anyone inside the lift, and accelerating along with it, would see the clocks keeping different rates of time. Hmm... I'll think about that. It would make sense if he didn't, but can you prove this?

The relative velocity/acceleration between the clocks and the person inside the lift is zero. The gravity is uniform. He would not see any difference in the clocks.

I'm not sure that's the right answer. Can you show mathematically how the observations of the free falling observer and the accelerating observer differ?

I'm not sure that's the right answer. Can you show mathematically how the observations of the free falling observer and the accelerating observer differ?

Why do you need math ? These are basic stuff.


Well, you ruled out gratvity already. The observer who jumps from the roof top will just float near the roof top. He is freefloating rather than freefalling. If there is any time dilation it might be because of the movement of the lift.



For the free floating observer :

The observer observes the lift moves with acceleration 'a'.

At the moment t (in his clock) he momentarily applies SR time dilation for the first & second clocks of the lift, v is the velocity of the lift clocks at that moment.

First clock's time, tc1 = t / sqrt (1 - v^2/c^2) ----> (1)
Second clock's time tc2 = t / sqrt (1 - v^2/c^2) -----> (2)

tc1 = tc2 because there is no relative velocity between the clocks in the lift.


at the moment t+dt , he observes/calculates time of the first & second clock that are now moving with different velocity v' = v+a*dt

First clock's time, tc1' = tc1+[dt / sqrt (1 - v'^2/c^2)] ----> (3)
Second clock's time, tc2' = tc2+[dt / sqrt (1 - v'^2/c^2)] ----> (4)

since tc1 = tc2 replace tc2 with tc1 in (4)
tc2' = tc1+[dt / sqrt (1 - v'^2/c^2)] ---->(5)

that is tc1' = tc2'

tc1'-tc1 = tc2' - tc2 ----> (6)

(6) means there is no time rate difference between clocks fixed in the lift.


The only time difference is between the observer's clock and the lift clocks.

since v <> v',

from (1) & (3)

tc1' - tc1 <> (t+dt) - t
tc1' - tc1 <> dt

from (6)

tc2' - tc2 <> dt

In short both the lift clocks show the same time but different from observer's own clock.

In total, There will not be any difference between the clocks fixed in the lift, for both freefloating observer and the observer inside the lift. Accleration due to motion does not cause the lift clocks to show any difference between them.

Well, I don't kmow about anybody, but I'm tired of jumping off roofs and being trapped in failed elevators. Let's try it this way.

I warn you, you'll need your wits about you, and is only very approximate..
Let's have 3 coordinate systems, K, K_i and K_a, each with a network of identical clocks. At t= 0 synchronise all clocks in K and K_a. Now let K_a accelerate away from K with uniform acceleration &gamma;.
Now let K_i be in uniform unnaccelerated motion relative to K with velocity = v. At a certain time synchronse all clocks in K_i and K_a. As before we consider the infitely small interval that v = &gamma;t.

Now, at t = 0, clocks in K_a were synchronous with each other and with clocks in K. As clocks in K_amove relative to K in the same way, relative to K they are synchronous with each other.

But, by the special theory, they are not synchronous with each other relative to K_i.
The simultaneity condition for events at points x₁ and x₂ in K_a is

t₁ - v(x₁)/c² = t₂ - v(x₂)/c² where v = &gamma;t. And if we set &xi; = x₂ - x₁ we have

t₂ = t₁(1 + &gamma;&xi;/c²), which by the principle of equivalence can be written
t₂ = t₁(1 + &Phi;/c²)

OK, that was a bit hard to follow. Here's another, equally approximate way of visualising it. Take a set of rectangular coordinates in a plane. Call the y-axis time (= t) and the x-axis....er....x, i.e. space in the x-direction.
Now plot the trajectory of A at rest, and call this A's world-line. Now the plane of simultaneity for A is simply any line that extends from a point on the t axis parallel to the x-axis. Let's assert that simultaneity for any body is defined by the line that intersects its world line at right angles.

Now draw the world-line of B moving with constant velocity relative to A, and again draw the plane of simultaneity at right angles to B's world line. We easily see that in both cases the slope of these planes is constant, i.e. takes the same value for all values of x.

Now draw the world line of C, moving with uniform acceleration relative to A. By the above definition of the plane of simultaneity, we can find no two values of x, no matter how close, at which the tangents to C's world line have the same slope. (EDIT: and of this applies equally to the tangent normals - did I need to say that?!) Hence, accepting that synchronicity of clocks and simultaneity mean exactly the same thing, clocks at different points in an accelerating body run at different rates.

These "space-time" diagrams are immensly powerful intuitive aids. You can for example, using the appropriate triangulations, easily extract the Lorentz tranform for time dilation and length contraction that Rosnet showed for Sciguy.

So here's a fun excersie. Use a space-time diagram to illustrate why the twin paradox occurs no matter what out and back trajectory one twin takes. (Hint: a frame is intertial iff a body moving uniformly in one coordinate system can be brought to rest by a single coordinate transformation)