Gravitational time dilation reason

[rogerk8]

Hi!

I am really bad at physics so bear with me...

I saw some very interesting 3 hour physics-marathon on TV the other day.

I have never gotten an actual clue on how things work relativisticly even though I've read a book by Brian Greene some years ago.

Since then I have tried to deduce the formula for time dilation and failed.

But seeing this TV-documentary made me actually deduce the equation for time dilation.

This equation being:

[tex]t0=t*\sqrt{1-\frac{v^2}{c^2}}[/tex]

Where t0 is the traveller.

I like this version better because it tells that same thing everybody is talking about i.e time stops when you are traveling at the speed of light.

But this was not my question.

My question is this:

What is the reason or theory behind gravity slowing down time?

I have searched the internet but all I can find is the formula

[tex]t0=t*\sqrt{1-\frac{r_{sch}}{r}}[/tex]

rsch being the schwartz-child radius of a black-hole equivalent for the planet.

And I have always been a curious guy so I ask myself why?

I can accept that the velocity of light is constant.

But what more do I have to accept?

That gravity stops time?

I am very interested in hearing some kind of physical explanation of why.

I know about the Doppler-effect. But that works on relative velocities (and not accelerations).

I do think there is some kind of connection here but I do not see it.

Thankful for any answer.

Roger
PS
Please excuse my bad english. I'm just a curious swede.

 

[WannabeNewton]

Do you know what the equivalence principle is?

 

[Jolb]

There is a lot of food for thought in these equations--especially those involving the Schwarzschild radius. But on a basic level, you can see why gravitational time dilation must occur without getting into advanced topics like the Schwarzschild radius by thinking about the following gedanken experiment (which I have stolen from Schutz's "A First Course in General Relativity").

First, assume there is a magical machine which can convert a photon of energy into some massive particle according to E=mc². Recall the energy of a photon: E=hv, where h is Planck's constant and v is the frequency of the photon. Further, assume we're staying close to the Earth where the potential energy of a massive particle is V=mgr, where m is the mass of the particle, r is its height above some reference point, and g is the gravitational acceleration in the region close to earth.

Now imagine we shoot a (massless) photon from the surface of the Earth (r=0) to the magical mass-energy converter suspended at a height r=R. If the photon's frequency v did not change between the surface of the Earth (r=0) and the magical converter, then we could generate energy out of nothing--starting with a photon with energy hv at the surface, if it arrives at the converter with energy hv, then we could convert it to a massive particle with E=hv=mc², then drop it down to the original spot, adding an additional energy mgR, so the final energy back at the surface is E'=hv+mgR. We could repeat the process and generate arbitrary amounts of energy out of nothing. This contradicts conservation of energy.

The answer is that the photon must have a different frequency when it reaches r=R due to gravitational time dilation. The decrease in the frequency is the manifestation of time dilation. This is the way GR avoids contradicting energy conservation.

Hope this argument for why gravitational time dilation must be there helps you. But there are other ways to see why it must happen.

 

[WannabeNewton]

Schutz got that argument from MTW by the way, so you aren't really guilty of stealing anything :)!

 

[rogerk8]

Thank you Jolb!

I'm sorry, I still don't understand.

Why would the photon gain mgH just by being dropped to the Earth surface?

This is how I see it:

Reaching H the photon energy is hf+mgH.

Dropping down to the surface of earth, the photon energy is hf (h=0).

Have I missed something?

Due to your nice example I however think I now understand that my initial thoughts about why gravity can bend light is correct.

The simple reason is that E=hf=mc^2 and this means that the photon can be given an equivalent mass and thereby be affected by the gravitational pull of large objects.

This insight also helps in stating the equations above.

This do however not help me understand why gravity affects time.

Roger

 

[rogerk8]

Do you know what the equivalence principle is?

No, please explain.

 

[WannabeNewton]

No, please explain.

See here: http://en.wikipedia.org/wiki/Gravitational_redshift#History

 

[Jolb]

Thank you Jolb!

I'm sorry, I still don't understand.

Why would the photon gain mgH just by being dropped to the Earth surface?

This is how I see it:

Reaching H the photon energy is hf+mgH.

Dropping down to the surface of earth, the photon energy is hf (h=0).

Have I missed something?

Hmmm, I think you have it a little mixed up. The experiment I described would go:
1) Photon initially at surface, which we will call altitude=0. Energy of photon=hv
2) Photon travels and ends up at an altitude=R. Here we make the assumption that E=hv. (This will be the assumption that leads to the contradiction.)
3) Photon enters the magical converter, which generates a particle at rest with mass m=E/c²=hv/c²
4) Particle of mass m falls freely to the surface. Its kinetic energy when it arrives at the surface is mgR. The total energy is mc² + mgR = (hv/c²)c²+mgR = hv+mgR > hv. This contradicts the conservation of energy.

Hence, at step 2), we made the wrong assumption. The correction that is needed has to do with the fact that the frequency v has been incorrectly assumed to be the same between steps 1) and 2). The frequency of the photon after step 2) should really be a smaller frequency, and the difference is due to gravitational time dilation.

Let me know if you understand now.

Do you know what the equivalence principle is?

No, please explain.

Roughly stated, the equivalence principle says that inertial mass is equivalent to gravitational mass. From a different perspective, one might think of the gravitational force as equivalent to the force a body feels when it is accelerated--that gravity is a fictitious force that arises from the curvature of spacetime. This concept is key to GR.

 

[RugbyEng]

I may be missing something, but initially the photon is "massless" with E=hv. It is then sent in the air into the magical energy to mass converter machine, where it is converted into a mass m=hv/c². Since it is no longer a photon it no longer has the energy hv, as it was converted to a mass, rather it has an energy mgR=hv (or it should be equal to this, provided the machine is 100% efficient). Using the above, hv/c²*gR=hv, thus, c^2=gR. This is not possible as c is constant, and g remains relatively constant. Therefore, v2 must decrease proportional to R, such that hv2/c²*gR=hv1 may hold.

Hmm, I hope that is accurate?

 

[WannabeNewton]

Here's the original argument from MTW (regarding energy conservation):
http://postimg.org/gallery/31ju2kf2/20673685/

 

[RugbyEng]

Here's the original argument from MTW (regarding energy conservation):
http://postimg.org/gallery/31ju2kf2/20673685/

oh crud, the next chapter was cut off...

 

[WannabeNewton]

Oh only the passages I posted are of relevance. You don't need to worry about the next section with regards to the above discussion.

 

[Jolb]

Seems like the gedanken experiment is giving people trouble. I'm at a loss on how to explain it more clearly.

The section in MTW is actually extremely terse, so here is the original section from Schutz, who goes a little more in-depth. MTW and Schutz are consistent in how they spell out the experiment, but my version changes the order around a little. (I start with a photon at the bottom, they start with a mass at the top. But it is the same argument.)

 

[rogerk8]

Seems like the gedanken experiment is giving people trouble. I'm at a loss on how to explain it more clearly.

The section in MTW is actually extremely terse, so here is the original section from Schutz, who goes a little more in-depth. MTW and Schutz are consistent in how they spell out the experiment, but my version changes the order around a little. (I start with a photon at the bottom, they start with a mass at the top. But it is the same argument.)

Hi Jolb!

You have been a little too fast...

You explained it so well that I now think I finally understand. The experiment anyway.

My interpretation:

As the photon "leaves" the gravitaional field the gain of mgR has to be reduced from hv due to conservation of energy.

Because otherwise the photon would gain energy simply by being reflected back from the moon, so to speak.

How close am I to the truth?

I do however still not understand how gravitation can affect time.

Unless the frequency of the photon itself can be related to time. Then strong gravitational fields obviously increses frequency and thereby reduces time.

How far fetched am I now?

Finally, your nice answer regarding the equivalence principle:

First, I read WannabeNewton's Wikipedia suggestion (History of Gravitational Redshift).

Guess what?

I read it twice but didn't understand a single thing :)

But your short explanation made me understand some things.

Most importantly that accelleration and gravitation are two sides of the same thing.

And I think the description of the curvature of space* is a beautiful description of how gravity of large planets affects moons.

Best Regards, Roger

 

[WannabeNewton]

Hi Roger! Let me try to explain to you the equivalence principle in a simple way. If you can get that down then the rest is just taking the kinematical time dilation between the bottom and top of the accelerating box in the accelerating box thought experiment explained in the wiki link and applying the equivalence principle to then state that there must be time dilation between different heights in a gravitational field.

Say we have an observer inside an elevator and say the observer is holding a small ball in his hand. First consider the scenario in which this elevator is at rest in a downwards uniform gravitational field ##\vec{g}##. If the observer were to drop the ball, he would see it fall to the floor of the elevator at a rate ##g##. Fair enough? Now consider the scenario in which the elevator is accelerating with ##\vec{a} = - \vec{g}##. If the observer were to drop the ball, he would again see it fall to the floor of the elevator at a rate ##g##. So operationally, the observer inside the elevator cannot discern between being at rest in a uniform gravitational field and accelerating in a direction opposite to that of the gravitational field. The two scenarios are physically equivalent.

 

[rogerk8]

I may be missing something, but initially the photon is "massless" with E=hv. It is then sent in the air into the magical energy to mass converter machine, where it is converted into a mass m=hv/c². Since it is no longer a photon it no longer has the energy hv, as it was converted to a mass, rather it has an energy mgR=hv (or it should be equal to this, provided the machine is 100% efficient). Using the above, hv/c²*gR=hv, thus, c^2=gR. This is not possible as c is constant, and g remains relatively constant. Therefore, v2 must decrease proportional to R, such that hv2/c²*gR=hv1 may hold.

Hmm, I hope that is accurate?

I find this very interesting. Hope you will get an expert's opinion soon.

Roger

 

[rogerk8]

Hi Roger! Let me try to explain to you the equivalence principle in a simple way. If you can get that down then the rest is just taking the kinematical time dilation between the bottom and top of the accelerating box in the accelerating box thought experiment explained in the wiki link and applying the equivalence principle to then state that there must be time dilation between different heights in a gravitational field.

Say we have an observer inside an elevator and say the observer is holding a small ball in his hand. First consider the scenario in which this elevator is at rest in a downwards uniform gravitational field ##\vec{g}##. If the observer were to drop the ball, he would see it fall to the floor of the elevator at a rate ##g##. Fair enough? Now consider the scenario in which the elevator is accelerating with ##\vec{a} = - \vec{g}##. If the observer were to drop the ball, he would again see it fall to the floor of the elevator at a rate ##g##. So operationally, the observer inside the elevator cannot discern between being at rest in a uniform gravitational field and accelerating in a direction opposite to that of the gravitational field. The two scenarios are physically equivalent.

Hi WannabeNewton!

Thank you for educating me!

I started to read it a third time but I couldn't even get over this initial statement:

"Once it became accepted that light is an electromagnetic wave, it was clear that the frequency of light should not change from place to place, since waves from a source with a fixed frequency keep the same frequency everywhere. One way around this conclusion would be if time itself were altered—if clocks at different points had different rates."

Please explain the bold part.

Regarding the rest, I give up.

The math is simply to hard.

But let's use what I do understand.

You pointed out that being in rest in a gravitational field is equivalent to accellerating even opposit that same gravitational field.

This I find very interesting.

My quest for understanding gravitational time dilation do however seem to just have begun :)

The question is, how do I get any wiser?

I really want to understand this.

But tensors, give me a break :)

Roger

 

[Jolb]

Hi Jolb!

You have been a little too fast...

You explained it so well that I now think I finally understand. The experiment anyway.

My interpretation:

As the photon "leaves" the gravitaional field the gain of mgR has to be reduced from hv due to conservation of energy.

That is absolutely true.

Because otherwise the photon would gain energy simply by being reflected back from the moon, so to speak.

How close am I to the truth?

Fairly close. Instead of reflecting it from the moon back to earth, (which would be completely symmetrical), to make the argument work, you would want to first convert it to a massive particle and then drop it back to earth. Converting between massive particles and massless photons is a key ingredient in the proof by contradiction. But more or less you have it.

I do however still not understand how gravitation can affect time.

Unless the frequency of the photon itself can be related to time.

It can be! This is another sort of crucial point that sometimes confuses people. If photons are generated on the surface with a frequency v=1 cycle per second (Hz), then if we sat a clock right next to the source of the photons, we'd see the second hand of the clock tick once for each cycle of the photon. The local correspondence between one tick per photon cycle is a local fact that cannot depend on reference frame.

But now we look at the photons from a higher altitude, and from there they are measured to have an energy hv-mgR and thus a new frequency v-mgR/h = 1-mgR/h, a slower frequency. The clock next to the source of the photons must be in synchronization with the photons! Thus looking down at the clock from a height R, we see the clock ticking once per 1/(1-mgR/h) seconds, which is longer than one second. So for a clock on the surface, then you will see one tick per second if you view it from the surface, but if you view the clock on the surface from a height R above it, you need to wait a slightly longer time for each tick. This is time dilation.

 

[WannabeNewton]

Roger let me be a little more quantitative. Consider a rocket ship of length ##h## at rest in the downwards uniform gravitational field ##g## of the Earth, an observer ##O## at the very top of the ship, and an observer ##O'## at the very bottom. Say ##O## has an ideal clock next to him and he uses it to send regularly time pulses of light down to ##O'##, with the regular pulses having an interval ##\Delta t_{O}## as measured by ##O##s clock. We want to know at what intervals ##\Delta t_{O'}## ##O'## receives these pulses as measured by an ideal clock next to him.

In the spirit of the above, consider now the situation in which the same ship is instead accelerating upwards in free space with magnitude ##g##. Now intuitively we can see that the light pulses sent by ##O## will reach ##O'## faster and faster because the ship is accelerating upwards so ##O'## will be catching up faster and faster with the light signals being sent down from the top of the ship by ##O##. Now, following the exposition in Hartle "Gravity: An Introduction to Einstein's General Relativity", assume for simplicity that the second order terms ##(V/c)^2## and ##(gh/c^2)^2## are negligible so that we can essentially use Newtonian mechanics. Then, choose an inertial frame that coincides with the bottom of the ship at ##t = 0## (call this the origin ##z = 0##) and in which the rocket accelerates upwards along the ##z##-axis of the inertial frame. The position of ##O'## will be given by ##z_{O'} = \frac{1}{2}gt^{2}## and the position of ##O## will be given by ##z_{O} = h + gt^{2}##.

Now imagine that at ##t = 0##, ##O## sends down a light pulse to ##O'## who receives it at a time ##t_1##. Then ##O## sends a second signal after the aforementioned interval ##\Delta t_{O}## and ##O'## receives it at a time ##t_1 + \Delta t_{O'}## because ##\Delta t_{O'}## represents the time between the first reception and the second reception as defined above hence the total time elapsed from the emission of the first pulse to the reception of the second is just the time ##t_1## from the emission to the first reception plus the time ##\Delta t_{O'}## from the first reception to the second. Now the first pulse travels a distance ##z_O (0) - z_{O'} (t_1) = ct_1## and the second pulse travels a distance ##z_O (\Delta t_O) - z_{O'}(t_1 + \Delta t_{O'}) = c(t_1 + \Delta t_{O'} - \Delta t_{O})## because ##\Delta t_{O} - t_1## gives us the time between the first reception and the second emission so ##\Delta t_{O'} - (\Delta t_{O} - t_1)## gives us the time between the second emission and second reception.

Plugging in our expressions for ##z_O## and ##z_{O'}## and discarding the negligible second order terms, we find that ##h - \frac{1}{2}gt_1^2 = ct^1, h - \frac{1}{2}gt_1^2 - gt_1 \Delta t_{O'} = t_1 + \Delta t_{O'} - \Delta t_{O}##. Solving this system of equations, we get ##\Delta t_{O'} = \Delta t_{O}(1 - \frac{gh}{c^{2}})##. Hence the observer at the bottom of the accelerating ship receives pulses from the observer at the top of the accelerating ship between intervals that are a fractional rate of the intervals between emissions by the observer at the top. Now the equivalence principle says that the observers inside the ship can equally well consider the ship as being at rest in the uniform gravitational field of the Earth in which case we interpret this as gravitational time dilation between different heights in the gravitational potential of the Earth's uniform gravitational field.

I hope this helps!

 

[rogerk8]

Roger let me be a little more quantitative. Consider a rocket ship of length ##h## at rest in the downwards uniform gravitational field ##g## of the Earth, an observer ##O## at the very top of the ship, and an observer ##O'## at the very bottom. Say ##O## has an ideal clock next to him and he uses it to send regularly time pulses of light down to ##O'##, with the regular pulses having an interval ##\Delta t_{O}## as measured by ##O##s clock. We want to know at what intervals ##\Delta t_{O'}## ##O'## receives these pulses as measured by an ideal clock next to him.

In the spirit of the above, consider now the situation in which the same ship is instead accelerating upwards in free space with magnitude ##g##. Now intuitively we can see that the light pulses sent by ##O## will reach ##O'## faster and faster because the ship is accelerating upwards so ##O'## will be catching up faster and faster with the light signals being sent down from the top of the ship by ##O##. Now, following the exposition in Hartle "Gravity: An Introduction to Einstein's General Relativity", assume for simplicity that the second order terms ##(V/c)^2## and ##(gh/c^2)^2## are negligible so that we can essentially use Newtonian mechanics. Then, choose an inertial frame that coincides with the bottom of the ship at ##t = 0## (call this the origin ##z = 0##) and in which the rocket accelerates upwards along the ##z##-axis of the inertial frame. The position of ##O'## will be given by ##z_{O'} = \frac{1}{2}gt^{2}## and the position of ##O## will be given by ##z_{O} = h + gt^{2}##.

Now imagine that at ##t = 0##, ##O## sends down a light pulse to ##O'## who receives it at a time ##t_1##. Then ##O## sends a second signal after the aforementioned interval ##\Delta t_{O}## and ##O'## receives it at a time ##t_1 + \Delta t_{O'}## because ##\Delta t_{O'}## represents the time between the first reception and the second reception as defined above hence the total time elapsed from the emission of the first pulse to the reception of the second is just the time ##t_1## from the emission to the first reception plus the time ##\Delta t_{O'}## from the first reception to the second. Now the first pulse travels a distance ##z_O (0) - z_{O'} (t_1) = ct_1## and the second pulse travels a distance ##z_O (\Delta t_O) - z_{O'}(t_1 + \Delta t_{O'}) = c(t_1 + \Delta t_{O'} - \Delta t_{O})## because ##\Delta t_{O} - t_1## gives us the time between the first reception and the second emission so ##\Delta t_{O'} - (\Delta t_{O} - t_1)## gives us the time between the second emission and second reception.

Plugging in our expressions for ##z_O## and ##z_{O'}## and discarding the negligible second order terms, we find that ##h - \frac{1}{2}gt_1^2 = ct^1, h - \frac{1}{2}gt_1^2 - gt_1 \Delta t_{O'} = t_1 + \Delta t_{O'} - \Delta t_{O}##. Solving this system of equations, we get ##\Delta t_{O'} = \Delta t_{O}(1 - \frac{gh}{c^{2}})##. Hence the observer at the bottom of the accelerating ship receives pulses from the observer at the top of the accelerating ship between intervals that are a fractional rate of the intervals between emissions by the observer at the top. Now the equivalence principle says that the observers inside the ship can equally well consider the ship as being at rest in the uniform gravitational field of the Earth in which case we interpret this as gravitational time dilation between different heights in the gravitational potential of the Earth's uniform gravitational field.

I hope this helps!

I think I've never felt so honored in my whole life!

Here I am, knowing extremely little about physics.

And yet, you put so much energy and effort into helping me understand.

It is amazing.

Thank you!

This first read did however leave several questions.

But the equations are fairly simple so if you give me a week I might be able to fully understand them.

Strike might, I will fully understand them. I am sure about it.

Otherwise, you will see me here asking more stupid questions :)

Take care!

Roger

 

[WannabeNewton]

Have fun Roger! By the way, I made a typo: the expression ##h - \frac{1}{2}gt_1^2 = ct^1## should read ##h - \frac{1}{2}gt_1^2 = ct_1##. Cheers.

 

[rogerk8]

Have fun Roger! By the way, I made a typo: the expression ##h - \frac{1}{2}gt_1^2 = ct^1## should read ##h - \frac{1}{2}gt_1^2 = ct_1##. Cheers.

I saw that ;)

Cheers to you too!

 

[WannabeNewton]

Oh and before I forget, the following expression ##h - \frac{1}{2}gt_1^2 - gt_1 \Delta t_{O'} = t_1 + \Delta t_{O'} - \Delta t_{O}## should actually be ##h - \frac{1}{2}gt_1^2 - gt_1 \Delta t_{O'} = c(t_1 + \Delta t_{O'} - \Delta t_{O})##. I forgot to tack on the ##c##, sorry about that.

 

[rogerk8]

That is absolutely true.


Fairly close. Instead of reflecting it from the moon back to earth, (which would be completely symmetrical), to make the argument work, you would want to first convert it to a massive particle and then drop it back to earth. Converting between massive particles and massless photons is a key ingredient in the proof by contradiction. But more or less you have it.


It can be! This is another sort of crucial point that sometimes confuses people. If photons are generated on the surface with a frequency v=1 cycle per second (Hz), then if we sat a clock right next to the source of the photons, we'd see the second hand of the clock tick once for each cycle of the photon. The local correspondence between one tick per photon cycle is a local fact that cannot depend on reference frame.

But now we look at the photons from a higher altitude, and from there they are measured to have an energy hv-mgR and thus a new frequency v-mgR/h = 1-mgR/h, a slower frequency. The clock next to the source of the photons must be in synchronization with the photons! Thus looking down at the clock from a height R, we see the clock ticking once per 1/(1-mgR/h) seconds, which is longer than one second. So for a clock on the surface, then you will see one tick per second if you view it from the surface, but if you view the clock on the surface from a height R above it, you need to wait a slightly longer time for each tick. This is time dilation.

You are a great teacher, Jolb!

I think I totally get it now.

Thanks!

To summarize:

The formulas and ideas needed for proving GTD:

1) E=mc^2
2) m=hv/c^2
3) V=mgR
4) c=constant
5) Conservation of energy

Of all these formulas, E=mc^2 is of course the neatest. And I do wonder about the reasoning behind that one too. Would be interesting to know.

One last stupid question:

Isn't the gravitational acceleration g in V above defined at the surface of Earth (r=3981 miles) where it even varies according to hight?

Finally, I find it very fascinating that the law of energy conservation is so vital in this fundamental case while the Big Bang is said to have happened out of nothing.

Roger

 

[Jolb]

You are a great teacher, Jolb!

I think I totally get it now.

Thanks!

To summarize:

The formulas and ideas needed for proving GTD:

1) E=mc^2
2) m=hv/c^2
3) V=mgR
4) c=constant
5) Conservation of energy

Of all these formulas, E=mc^2 is of course the neatest. And I do wonder about the reasoning behind that one too. Would be interesting to know.

One last stupid question:

Isn't the gravitational acceleration g in V above defined at the surface of Earth (r=3981 miles) where it even varies according to hight?

1), 4), and 5) are fundamental equations in general relativity. 1) E=mc² is a result of special relativity, and it can be derived by demanding that the energy of a massive particle must be lorentz invariant. (For example, if we used the Newtonian expression for kinetic energy, K=mw²/2, this clearly depends on reference frame since the particle's velocity w is frame-dependent. (Sorry for the w but I've already used v for frequency--which if I weren't lazy is usually the greek letter nu.) The real form of the equation derived in this way is E=(p²c²+m²c⁴)^1/2, which is a Lorentz invariant quantity, and the expression reduces to 1) for a particle at rest.]

2) is not a fundamental equation but describes the magical photon-to-massive-particle converter in the gedanken experiment. Such a device is in principle possible in relativity, so there is no problem there.

3) is a simplifying assumption that I used, and you are right to point out that it is an inexact approximation to the gravitational potential close to the surface of a planet. In other words, V=mgr is an approximation to order r, and it only holds when r<<R, the radius of the planet. You can use any other realistic form of V for this thought experiment, such as the more general V=-GMm/r, where G is the Newtonian gravitational constant, M is the total mass of the planet, and r is the distance from the planet's core. The thought experiment would work out just the same, but the algebra is slightly more involved.

Finally, I find it very fascinating that the law of energy conservation is so vital in this fundamental case while the Big Bang is said to have happened out of nothing.

Roger

One interesting thing in GR is that there is no "time" prior to the Big Bang. So there really isn't anything "before" the big bang that we could compare to the energy of the universe after the big bang.

 

[rogerk8]

1), 4), and 5) are fundamental equations in general relativity. 1) E=mc² is a result of special relativity, and it can be derived by demanding that the energy of a massive particle must be lorentz invariant. (For example, if we used the Newtonian expression for kinetic energy, K=mw²/2, this clearly depends on reference frame since the particle's velocity w is frame-dependent. (Sorry for the w but I've already used v for frequency--which if I weren't lazy is usually the greek letter nu.) The real form of the equation derived in this way is E=(p²c²+m²c⁴)^1/2, which is a Lorentz invariant quantity, and the expression reduces to 1) for a particle at rest.]

2) is not a fundamental equation but describes the magical photon-to-massive-particle converter in the gedanken experiment. Such a device is in principle possible in relativity, so there is no problem there.

3) is a simplifying assumption that I used, and you are right to point out that it is an inexact approximation to the gravitational potential close to the surface of a planet. In other words, V=mgr is an approximation to order r, and it only holds when r<<R, the radius of the planet. You can use any other realistic form of V for this thought experiment, such as the more general V=-GMm/r, where G is the Newtonian gravitational constant, M is the total mass of the planet, and r is the distance from the planet's core. The thought experiment would work out just the same, but the algebra is slightly more involved.
One interesting thing in GR is that there is no "time" prior to the Big Bang. So there really isn't anything "before" the big bang that we could compare to the energy of the universe after the big bang.

Hi Jolb!

Very interesting but could you please try to explain the bold part again?

I am confused.

For me R>>r would mean that g=MG/r^2 has dimnished to almost zero at r=R, so to speak.

R being the distance to observer and r being the planet radious.

Regarding your BB opinion, I think it sounds a bit too convenient.

Roger

 

[Jolb]

Yes, that standard reply to "why does the Big Bang not violate conservation of energy?" leaves many people, me included, unsatisfied. Then again I'm a quantum guy and I usually view GR with a grain of salt.

As to the form of the gravitational potential, I'll just show you a derivation (which is little more than Taylor series.) [Please, mods, don't give me an infraction for too much help!]

From Newtonian gravity, we know that the gravitational potential energy of a massive particle outside a spherical body, e.g. a planet, is given by V=-GMm/r, where M is the mass of the planet, m is the mass of the particle, r is the distance from the core of the planet to the particle, and G is the Newtonian gravitational constant. At the surface of the planet, r=R, we have V=-GMm/R. If we taylor expand around r=R, we get: [tex]
V=-\frac{GMm}{r}=\sum_{n=0}^\infty\frac{(r-R)^n}{n!} \frac{d^nV}{dr^n} |_{r=R}=V(R)+(r-R)\frac{dV}{dr} |_R+\frac{(r-R)^2}{2}\frac{d^2V}{dr^2}|_R + ...[/tex]
[See http://en.wikipedia.org/wiki/Taylor_expansion for the details of Taylor expansion, which is just pure mathematics]
If we go ahead and take the derivatives:
[tex]
\frac{dV}{dr} |_R=\frac{d}{dr} \left [ \frac{-GMm}{r} \right]_R=\frac{GMm}{r^2} |_R=\frac{GMm}{R^2}[/tex]
similarly,
[tex]\frac{d^2V}{dr^2} |_R=\frac{d}{dr} \left [ \frac{GMm}{r^2}\right ]_R=-2\frac{GMm}{R^3}[/tex]
Plugging these into the taylor expansion:
[tex]V=-\frac{GMm}{R}+(r-R)\frac{GMm}{R^2}+\frac{(r-R)^2}{2}\left(- 2\frac{GMm}{R^3} \right) + ...[/tex]
[tex]=-\frac{GMm}{R}+\frac{r-R}{R}\left (\frac{GMm}{R} \right )-\left (\frac{r-R}{R} \right )^2\left(\frac{GMm}{R} \right) + ...[/tex]

The first term is a constant, and for a potential energy function, we are always allowed to throw away constant contributions to the energy. Now since we are in the regime r-R<<R, this implies (r-R)/R << 1, and thus [(r-R)/R]²<<(r-R)/R. So the third term is negligible compared to the second. The fourth term will be proportional to [(r-R)/R]^3, the fifth proportional to [(r-R)/R]^4, and so forth, so all the rest of the terms are even more negligible. We throw away all the negligible terms to give a good approximation in the regime r-R<<R. So for an approximate equation we can use the symbol ≈ to indicate non-exact equality.

Thus we are left with V≈(r-R)/R * (GMm/R) ≈ (r-R)*(GMm/R²). Now, if we set r-R=a, the altitude, and if we call the quantity (GM/R²)=g, then we have V≈mga, as before.

If you stare at this derivation for a little while your confusion should go away. This derivation really shows how all the variables (r, R, and a) are related, and hopefully you can identify places in the above posts where r should actually be a, the altitude above the surface. (Anytime I used the formula V=mgr, that should really be V=mga to be notationally consistent with the above derivation.)

Hopefully there aren't too many typos in that. (I did that from scratch as a little exercise, so there may be errors, but you can find the same reasoning in any Mechanics textbook.) But taylor expansions which lead to approximations are really the bread and butter of physicists.

 

[rogerk8]

Yes, that standard reply to "why does the Big Bang not violate conservation of energy?" leaves many people, me included, unsatisfied. Then again I'm a quantum guy and I usually view GR with a grain of salt.

As to the form of the gravitational potential, I'll just show you a derivation (which is little more than Taylor series.) [Please, mods, don't give me an infraction for too much help!]

From Newtonian gravity, we know that the gravitational potential energy of a massive particle outside a spherical body, e.g. a planet, is given by V=-GMm/r, where M is the mass of the planet, m is the mass of the particle, r is the distance from the core of the planet to the particle, and G is the Newtonian gravitational constant. At the surface of the planet, r=R, we have V=-GMm/R. If we taylor expand around r=R, we get: [tex]
V=-\frac{GMm}{r}=\sum_{n=0}^\infty\frac{(r-R)^n}{n!} \frac{d^nV}{dr^n} |_{r=R}=V(R)+(r-R)\frac{dV}{dr} |_R+\frac{(r-R)^2}{2}\frac{d^2V}{dr^2}|_R + ...[/tex]
[See http://en.wikipedia.org/wiki/Taylor_expansion for the details of Taylor expansion, which is just pure mathematics]
If we go ahead and take the derivatives:
[tex]
\frac{dV}{dr} |_R=\frac{d}{dr} \left [ \frac{-GMm}{r} \right]_R=\frac{GMm}{r^2} |_R=\frac{GMm}{R^2}[/tex]
similarly,
[tex]\frac{dV}{dr} |_R=\frac{d}{dr} \left [ \frac{GMm}{r^2}\right ]_R=-2\frac{GMm}{R^3}[/tex]
Plugging these into the taylor expansion:
[tex]V=-\frac{GMm}{R}+(r-R)\frac{GMm}{R^2}+\frac{(r-R)^2}{2}\left(- 2\frac{GMm}{R^3} \right) + ...[/tex]
[tex]=-\frac{GMm}{R}+\frac{r-R}{R}\left (\frac{GMm}{R} \right )-\left (\frac{r-R}{R} \right )^2\left(\frac{GMm}{R} \right) + ...[/tex]

The first term is a constant, and for a potential energy function, we are always allowed to throw away constant contributions to the energy. Now since we are in the regime r-R<<R, this implies (r-R)/R << 1, and thus [(r-R)/R]²<<(r-R)/R. So the third term is negligible compared to the second. The fourth term will be proportional to [(r-R)/R]^3, the fifth proportional to [(r-R)/R]^4, and so forth, so all the rest of the terms are even more negligible. We throw away all the negligible terms to give a good approximation in the regime r-R<<R. So for an approximate equation we can use the symbol ≈ to indicate non-exact equality.

Thus we are left with V≈(r-R)/R * (GMm/R) ≈ (r-R)*(GMm/R²). Now, if we set r-R=a, the altitude, and if we call the quantity (GM/R²)=g, then we have V≈mga, as before.

If you stare at this derivation for a little while your confusion should go away. This derivation really shows how all the variables (r, R, and h) are related, and hopefully you can identify places in the above posts where r should actually be a, the altitude above the surface. (Anytime I used the formula V=mgr, that should really be V=mga to be notationally consistent with the above derivation.)

Hopefully there aren't too many typos in that. (I did that from scratch as a little exercise, so there may be errors, but you can find the same reasoning in any Mechanics textbook.) But taylor expansions which lead to approximations are really the bread and butter of physicists.

Jesus Jolb, this is the second time I've felt really honored at this nice forum!

Thank you very much for your time and effort!

First, I'm surpriced to hear your doubts.

As well as releaved, I might add.

Because I wanted it to be said without violating the rules.

Back to topic.

You might be surprised to hear that I actually understood the Taylor expansion.

That is one of the few mathematical tricks I know.

I do not know if this is true but I think calculators use Taylor series in their evaluation of functions such as exp(x) for instance because it then comes down to only the four simple and basic mathematical functions, namely addition, substraction, division and multiplication.

Please regard this as a parenthesis only :)

Ok, in this case R seems to be the planet radious, right?

So if r-R=a=the altitude and (GM/R^2)=g then V=mga (where g only depends on R) and my confusion has just vanished :D

Roger

 

[Jolb]

Jesus Jolb, this is the second time I've felt really honored at this nice forum!

Thank you very much for your time and effort!

You're welcome!

I do not know if this is true but I think calculators use Taylor series in their evaluation of functions such as exp(x) for instance because it then comes down to only the four simple and basic mathematical functions, namely addition, substraction, division and multiplication.

Yes, I believe most handheld calculators do use Taylor series in evaluating functions like exp(x), but electrical engineers get more and more creative with designing hardware that can perform calculations more quickly than any software loop, so perhaps there are certain calculators which can exponentiate on a hardware level.

But you might be interested in a very nice connection between taylor series and the exponential function. If you remember the two facts exp(0)=1 and d/dx exp(x) = exp(x), you can easily plug in every term in its taylor series to get the beautiful expression
[tex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1+x+\frac{x^2}{2} + \frac{x^3}{3!}+\frac{x^4}{4!}+...[/tex]
And that expression is what many mathematicians take to be the definition of the exponential function. You can see nice things happen when you take the derivatives and integrals of that expression.

Ok, in this case R seems to be the planet radious, right?

So if r-R=a=the altitude and (GM/R^2)=g then V=mga (where g only depends on R) and my confusion has just vanished :D

Exactly. Except don't forget that the total mass of the planet M also goes into determining g at the surface.

Also, in your last post you quoted me before I had a chance to fix a typo. I somehow put "h" in where there should have been an "a" in the sentence "This derivation really shows how all the variables (r, R, and [STRIKE]h[/STRIKE]) are related..."

Edit: Found another error. I have dV/dr instead of d²V/dr² in the line where I take the second derivative. I'll edit it in the original post but of course your quote of that post will still have the error.

 

[rogerk8]

Hi Jolb!

Regarding your late edit...

I actually stared at that part for a while but then I thought I got it.

Obviously I didn't :)

Roger

 

[rogerk8]

WannabeNewton,

I got stuck early in your explanation i.e at:

The position of ##O'## will be given by ##z_{O'} = \frac{1}{2}gt^{2}## and the position of ##O## will be given by ##z_{O} = h + gt^{2}##.

Please explain this like I have no clue about physics at all.

If something is free-falling, does the length it falls equal 1/2gt^2. If so, why?

And why is it gt^2 for O?

Roger

 

[WannabeNewton]

Hi roger! For an object under the force ##mg##, the path in general is given by ##z(t) = z_0 + v_0 t + \frac{1}{2}gt^2## where ##z_0## is the initial position and ##v_0## is the initial velocity. This is gotten simply by solving Newton's 2nd law for ## m\ddot{x} = F = mg##. ##O'##s initial position is at the bottom so he has ##z_0 = 0## and he starts out at rest so ##v_0 = 0##. ##O## is at the top so his initial position is ##h## (the height of the rocket) and he also starts out at rest so ##v_0 = 0##.

 

[rogerk8]

Hi roger! For an object under the force ##mg##, the path in general is given by ##z(t) = z_0 + v_0 t + \frac{1}{2}gt^2## where ##z_0## is the initial position and ##v_0## is the initial velocity. This is gotten simply by solving Newton's 2nd law for ## m\ddot{x} = F = mg##. ##O'##s initial position is at the bottom so he has ##z_0 = 0## and he starts out at rest so ##v_0 = 0##. ##O## is at the top so his initial position is ##h## (the height of the rocket) and he also starts out at rest so ##v_0 = 0##.

Hi WannabeNewton!

Would you mind solving the differential equation for me also?

I think I can see that if you integrate x'' once you'll get vt and if you integrate it twice you'll get 1/2gt^2.

But it was such a long time since I worked with differential equations so I do not remember and I cannot see it clearly.

Hope you don't think I am lazy for asking this.

Roger

 

[rogerk8]

Hi WannabeNewton!

Solving...

[tex]mx''=F=mg[/tex]


[tex]x''=F/m=g[/tex]

[tex]x'=F/m(t+A)[/tex]

[tex]x=F/m(t^2/2+At+B)[/tex]

[tex]x(0)=F/m*B=x_0[/tex]

[tex]x'(o)=F/m*A=v_0[/tex]

Which leads to

[tex]x(t)=x_0+v_0t+1/2gt^2[/tex]

Thank you for not solving it for me!

I will not be that lazy again :)

Roger

 

[WannabeNewton]

Yep that works fine :)