Thread: Photon clock entering lower grav. pot.: What happens ?

In kinetic time dilatation the photon clock runs slow due to relative speed, because for the outside observer the path is observed as longer while the photon travels at constant speed c. So time must slow down.

An atomic clock has been proven to run slow when entering a gravitational well,
but how are we to understand the slowing down of the photon clock for this case ?

Originally Posted by Noa Drake

In kinetic time dilatation the photon clock runs slow due to relative speed, because for the outside observer the path is observed as longer while the photon travels at constant speed c. So time must slow down.

An atomic clock has been proven to run slow when entering a gravitational well,
but how are we to understand the slowing down of the photon clock for this case ?

In the case of a light clock, the deeper the clock is in the gravity well, the longer the path between the mirrors, it is "as if" the gravity acts as a strainer, "lengthening" the clock. So, the clock will tick slower, by the exact amount predicted by GR.
The fascinating thing is that light moving "up", against the gravitational field will take longer than light "falling down", with the gravitational field. But when you add the two halves of the light clock period, you still get a longer period, proportional to the gravitational acceleration.

How would the outside observer be able to observe this ? Standing on the earth watching a radially incoming light clock, with the photon travelling up and down, parallel to his line of sight ? (This would be the exact same clock as set up for kinetic time dilatation).

Originally Posted by Noa Drake

How would the outside observer be able to observe this ? Standing on the earth watching a radially incoming light clock, with the photon travelling up and down, parallel to his line of sight ? (This would be the exact same clock as set up for kinetic time dilatation).

He wouldn't. Al the clocks next to him (atomic, mechanical, etc) would be slowed down by the same exact factor. The only way to figure out that the clocks are running slower is by comparison with a clock "up" in space, like having a GPS satellite transmit its time stamps.

So for gravitational time dilatation it is not necessary for the outside observer to be in the same location as his reference clock ?

Originally Posted by Noa Drake

So for gravitational time dilatation it is not necessary for the outside observer to be in the same location as his reference clock ?

You have it backwards, one more (and last) time, an observer next to the clock will not observe any time dilation. Gravitational or kinematic.

I am not talking about any observers next to the clock.

It is about the observer on earth, standing next to his reference clock, as indicated, looking up ant the incoming photon clock.

He does not observe any path for the incoming photon of the clock lenghtened, as you indicated.
(Whereas that was the basis of the calculation for the case of kinetic time dilatation.)

So my question was that apparently that is not a recquirement in gravitational time dilatation, based on your answers.

Originally Posted by Noa Drake

I am not talking about any observers next to the clock.

It is about the observer on earth, standing next to his reference clock, as indicated, looking up ant the incoming photon clock.

What will such an observer measure? The "incoming clock" is encased.

He does not observe any path for the incoming photon of the clock lenghtened, as you indicated.

He cannot observe anything but he can use the predictions of the theory of relativity to calculate the period of the "incoming clock".

So my question was that apparently that is not a recquirement in gravitational time dilatation, based on your answers.

False, you got it backwards. Not surprising.

It is really simple: a clock in a (uniform) gravitational field is equivalent to a clock accelerated at (constant) acceleration according to the Equivalence Principle.
When light leaves the "floor" of the clock going towards the mirror at the "top", the light path, from the perspective of an inertial observer follows the equation:



The above degree two equation has the solution

In the reverse direction, the equation is:



with the solution:



The clock period is :



It is easy to show that , the period of a clock in zero gravitational field (located in the inertial frame). I leave this as an exercise for you.

One more thing: before you start making your standard wild claims, the above has been verified experimentally, by Vessot and by the experiments ran on Gravity Probe A.

I have always claimed to agree on the formulas of gravitational and kinetic time dilatation, and still do.

(The road to get there is another matter, that i will not discuss in this section.)


I have made no claims in this section, you gave answers. But judging from your answers, and specifically,

"He cannot observe anything but he can use the predictions of the theory of relativity to calculate the period of the "incoming clock".
"The only way to figure out that the clocks are running slower is by comparison with a clock "up" in space, like having a GPS satellite transmit its time stamps."

it is clear that for gravitational time dilatation, the principle of observer dependent time is not applied to obtain the formulas.


Yet both formulas have the same format : 1/ ((1 - (v/c)²) and 1/ (1 - (2GM/rc²))
>>>> v for kinetic t.d. and 2GM/r for gravitational t.d. as the changing factors, resulting in the exact same corellation.


I am not aiming anything at you personally, you are in this case the defender of a theory and that is just fine by me, i just try to make sense of how it is constructed.

Originally Posted by Noa Drake

it is clear that for gravitational time dilatation, the principle of observer dependent time is not applied to obtain the formulas.

False, I explained very clearly , with math, that the period of the light clock is the one predicted by the inertial observer, not by the observer traveling with the clock.

It is really simple: a clock in a (uniform) gravitational field is equivalent to a clock accelerated at (constant) acceleration according to the Equivalence Principle.
When light leaves the "floor" of the clock going towards the mirror at the "top", the light path, from the perspective of an inertial observer follows the equation:



The above degree two equation has the solution

In the reverse direction, the equation is:



with the solution:



The clock period is :



It is easy to show that , the period of a clock in zero gravitational field (located in the inertial frame). I leave this as an exercise for you.

One more thing: before you start making your standard wild claims, the above has been verified experimentally, by Vessot and by the experiments ran on Gravity Probe A.

*I never spoke anywhere of an observer travelling with the clock.
*I made no wild claims.

*And i agree of course with :
"a clock in a (uniform) gravitational field is equivalent to a clock accelerated at (constant) acceleration according to the Equivalence Principle."




*Can you clarify or visualise the perspective of the inertial observer ?

From this : When light leaves the "floor" of the clock going towards the mirror at the "top", the light path, from the perspective of an inertial observer follows the equation:

As for the experiments performed: Vessot, Gravity Probe) : hydrogen masers :

"The interaction of the electron and proton in the hydrogen atom generates a microwave signal "

So this is proof, solid proof for this type of clock, no problem.
And i have no doubt about muons arriving later and classical clocks going slow as well.
(And i also agree on the speed of light staying constant.)

But we are talking about a light clock, that was my specific angle in post 1.
I focused on this type of clock because it serves as the foundation from which time dilation was derived by Einstein.

Originally Posted by Noa Drake

*I never spoke anywhere of an observer travelling with the clock.
*I made no wild claims.

*And i agree of course with :
"a clock in a (uniform) gravitational field is equivalent to a clock accelerated at (constant) acceleration according to the Equivalence Principle."




*Can you clarify or visualise the perspective of the inertial observer ?

From this : When light leaves the "floor" of the clock going towards the mirror at the "top", the light path, from the perspective of an inertial observer follows the equation:

I already explained that in the test: while the beam moves "up" by , the "top" of the clock also moves up by so, the beam must cover a distance
In the opposite direction, things are different, the "floor" of the clock moves towards the incoming beam, such that the distance covered needs only be . This is basic kinematics.

" while the beam moves "up", the "top" of the clock also moves up "

" In the opposite direction, things are different, the "floor" of the clock moves towards the incoming beam "


> I don't see how this happens, can you explain why this is the case ?

Originally Posted by Noa Drake

" while the beam moves "up", the "top" of the clock also moves up "

" In the opposite direction, things are different, the "floor" of the clock moves towards the incoming beam "


> I don't see how this happens, can you explain why this is the case ?

OK,

I'll try to reduce the situation to a simpler one, no accelerated motion, no GR.
Imagine that you set the light clock horizontal and that it moves at speed v wrt to an observer. The clock moves away from the observer, in the direction of the positive x axis. When the beam goes L to R, the equation is:





When the beam bounces back, R to L, the equation becomes:






The period of the moving clock is the sum of the two trips:



But wait! The length of the clock is contracted , due to motion, so
Therefore:



which is exactly the SR formula for time dilation, since is the period of the clock at rest wrt the observer. Hope this helped. If it didn't, nothing else will.

Originally Posted by Noa Drake

Yet both formulas have the same format : 1/ ((1 - (v/c)²) and 1/ (1 - (2GM/rc²))

Just a quick note to say that the above is true only in the case of the Schwarzschild metric - other solutions ( e.g. if the central body rotates, or isn't perfectly spherical etc etc ) will yield different formulas for gravitational time dilation. In general, the time dilation factor is the square root of the (0,0) component of the metric tensor.

Ok, thank you Howard and Markus for your responses.

Originally Posted by Noa Drake

Ok, thank you Howard and Markus for your responses.

You are welcome.