Jump to content

The fall of the apple and the LC oscillatory circuit in the reference frame of the remote observer.


SergUpstart
 Share

Recommended Posts

Let's do a thought experiment. A tree grows on planet X, an apple falls from it from a height H. There is an observer next to the tree,
and in his frame of reference, the apple will fall in time

t2h.jpg

Let another observer from outer space also observe the fall of an apple through a telescope. Due to the fact that the second observer is < br /> in a weaker gravity, in its frame of reference, the time of the apple's fall will be slowed down by a factor of k, and the height of H will also be increased by a factor of k.
gthh.jpg
 
From this it can be seen that in the reference frame of a remote observer, the acceleration of gravity on the surface of planet X should be k times less.
This change also corresponds to the dimension of the acceleration of free fall, m/s^2. But on the other hand
 
gg.jpg
 
where M and R are the mass and radius of planet X, respectively. In the reference frame of a remote observer, the radius of planet X will also increase by a factor k,
and the acceleration of gravity on the surface of planet X g' will be
 
ggg.jpg



This shows that in order for the acceleration of free fall on the surface of planet X in sesame starting a remote observer g' would be k times
less than it is in the system of reference of the observer insider, requires that the gravitational constant G in the reference frame of an external observer would also
k times greater than in the reference system of observer-insider. Otherwise, in the reference frame of the remote observer, the acceleration of gravity g' will not be in k, but in k^2
times less than it is in the frame of reference of the insider observer.


This simple thought experiment proves that in the reference frame of a remote observer, the gravitational constant must change back
proportional to the change in the distance/time scale. 

Next, we consider the LC oscillatory circuit. The capacity of the capacitor in the frame of reference of the insider observer is calculated by the formula.
capocity.jpg

First, let's see how the capacitance of the capacitor will change in the reference frame of the remote observer, if we assume that the dielectric constant on changes.
In the reference frame of an external observer, the area of the capacitor plates will increase by k^2 times, and the distance between the plates will also increase by k times,
thus, the capacitance of the capacitor will increase by k times.
Next, let's see how the inductance of the coil will change in the reference system of the remote observer, which is in the reference system of the insider observer
it is calculated by the formula.
indu.jpg

First, let's see how the capacitance of the capacitor will change in the reference frame of the remote observer, if we assume that the dielectric constant on changes.
In the reference frame of an external observer, the area of the capacitor plates will increase by k^2 times, and the distance between the plates will also increase by k times,
thus, the capacitance of the capacitor will increase by k times.
Next, let's see how the inductance of the coil will change in the reference system of the remote observer, which is in the reference system of the insider observer
it is calculated by the formula.
tlcc.jpg

We see that the oscillation period increases by k times, as it should be.
But the dimension of the electric constant F/m, so that in the reference frame of a remote observer, it should decrease by k times,
then, in order for the speed of light to remain constant, the magnetic constant must increase by k times.
Then, in the reference frame of the remote observer, the capacitance of the capacitor will not change, but the inductance of the coil will increase by k^2 times.
Thus, the oscillation period of the oscillating circuit will also increase by k times in this case.


To find out which of these two options is correct, we will introduce an active resistance into the oscillatory circuit
so that the fluctuations in it become attenuated.
Let's see how the resistance value changes when switching to the reference system of a remote observer. Resistance is the ratio
the difference of electrical potentials to the current strength in the conductor. The dimension of the electric potential is Joule/Coulomb.
The joule does not change when the time/distance scale changes synchronously, and the Сoulomb is not related to the time/distance scales at all,
therefore, the difference of electrical potentials in the reference system of the remote observer will be the same as in the reference system
an insider's observer. The strength of the electric current in the conductor is equal to the ratio of the charge that has passed through the cross-section of the conductor to the time for which this charge passed. The charge in the reference frame of the external observer will be the same as in the reference frame of the insider observer but the time will increase by k times.
Thus, the resistance in the reference frame of the remote observer will be k times greater than in the reference frame of the insider-observer.

In order for both the insider observer and the remote observer to see the same picture of damped oscillations on different time scales,
it is required that the Q-factor of the oscillatory circuit in the reference systems of both observers would be the same.
(Q-factor is a parameter of the oscillatory system that determines the resonance width and characterizes,
how many times are the energy reserves in the system greater than the energy losses during the phase change by 1 radian.)
qflcc.jpg

In the first variant, when the dielectric and magnetic constants are considered unchanged, in the reference frame of the remote observer, the Q-factor
the oscillating circuit will be
qflcc1.jpg


 
And in the second variant, when the dielectric and magnetic constants change in different directions by k times, in the reference frame of the remote observer
the Q-factor the oscillating circuit will be
qflcc2.jpg


 
Thus, this thought experiment with an oscillatory circuit shows that in the reference frame of a remote observer, the dielectric and magnetic constants must change in different directions in accordance with the change in the time/distance scale. Then it is not difficult to show
that the Planck constant will also change in the reference frame of the remote observer, since it is equal to
hhhh.jpg

 

For those who still do not believe that the numerical value of the gravitational constant in the reference frame of a remote observer changes back
in proportion to the change in the time/distance scale, I propose to solve the following problem.
On a planet with a mass of 5.97*10 ^ 24 kg, a radius of 6317000 m, an apple falls from a tree from a height of 5 m in 1 second, calculate the value of the gravitational  constant G. And then calculate the value of the gravitational constant G' in the reference frame of the remote observer, in which the apple falls for 1.1 seconds from a height of 5.5 m and the radius of the planet is 6948700, respectively, and compare the obtained values of G and G'.
 
applf.jpg

 

Link to comment
Share on other sites

46 minutes ago, SergUpstart said:
Let's do a thought experiment. A tree grows on planet X, an apple falls from it from a height H. There is an observer next to the tree,
and in his frame of reference, the apple will fall in time
 
t2h.jpg
 
Let another observer from outer space also observe the fall of an apple through a telescope. Due to the fact that the second observer is < br /> in a weaker gravity, in its frame of reference, the time of the apple's fall will be slowed down by a factor of k, and the height of H will also be increased by a factor of k.
gthh.jpg
 
From this it can be seen that in the reference frame of a remote observer, the acceleration of gravity on the surface of planet X should be k times less.
This change also corresponds to the dimension of the acceleration of free fall, m/s^2. But on the other hand
 
gg.jpg
 
where M and R are the mass and radius of planet X, respectively. In the reference frame of a remote observer, the radius of planet X will also increase by a factor k,
and the acceleration of gravity on the surface of planet X g' will be
 
ggg.jpg



This shows that in order for the acceleration of free fall on the surface of planet X in sesame starting a remote observer g' would be k times
less than it is in the system of reference of the observer insider, requires that the gravitational constant G in the reference frame of an external observer would also
k times greater than in the reference system of observer-insider. Otherwise, in the reference frame of the remote observer, the acceleration of gravity g' will not be in k, but in k^2
times less than it is in the frame of reference of the insider observer.


This simple thought experiment proves that in the reference frame of a remote observer, the gravitational constant must change back
proportional to the change in the distance/time scale. 

Next, we consider the LC oscillatory circuit. The capacity of the capacitor in the frame of reference of the insider observer is calculated by the formula.
capocity.jpg
 
First, let's see how the capacitance of the capacitor will change in the reference frame of the remote observer, if we assume that the dielectric constant on changes.
In the reference frame of an external observer, the area of the capacitor plates will increase by k^2 times, and the distance between the plates will also increase by k times,
thus, the capacitance of the capacitor will increase by k times.
Next, let's see how the inductance of the coil will change in the reference system of the remote observer, which is in the reference system of the insider observer
it is calculated by the formula.
indu.jpg
 
First, let's see how the capacitance of the capacitor will change in the reference frame of the remote observer, if we assume that the dielectric constant on changes.
In the reference frame of an external observer, the area of the capacitor plates will increase by k^2 times, and the distance between the plates will also increase by k times,
thus, the capacitance of the capacitor will increase by k times.
Next, let's see how the inductance of the coil will change in the reference system of the remote observer, which is in the reference system of the insider observer
it is calculated by the formula.
tlcc.jpg
 
We see that the oscillation period increases by k times, as it should be.
But the dimension of the electric constant F/m, so that in the reference frame of a remote observer, it should decrease by k times,
then, in order for the speed of light to remain constant, the magnetic constant must increase by k times.
Then, in the reference frame of the remote observer, the capacitance of the capacitor will not change, but the inductance of the coil will increase by k^2 times.
Thus, the oscillation period of the oscillating circuit will also increase by k times in this case.


To find out which of these two options is correct, we will introduce an active resistance into the oscillatory circuit
so that the fluctuations in it become attenuated.
Let's see how the resistance value changes when switching to the reference system of a remote observer. Resistance is the ratio
the difference of electrical potentials to the current strength in the conductor. The dimension of the electric potential is Joule/Coulomb.
The joule does not change when the time/distance scale changes synchronously, and the Сoulomb is not related to the time/distance scales at all,
therefore, the difference of electrical potentials in the reference system of the remote observer will be the same as in the reference system
an insider's observer. The strength of the electric current in the conductor is equal to the ratio of the charge that has passed through the cross-section of the conductor to the time for which this charge passed. The charge in the reference frame of the external observer will be the same as in the reference frame of the insider observer but the time will increase by k times.
Thus, the resistance in the reference frame of the remote observer will be k times greater than in the reference frame of the insider-observer.

In order for both the insider observer and the remote observer to see the same picture of damped oscillations on different time scales,
it is required that the Q-factor of the oscillatory circuit in the reference systems of both observers would be the same.
(Q-factor is a parameter of the oscillatory system that determines the resonance width and characterizes,
how many times are the energy reserves in the system greater than the energy losses during the phase change by 1 radian.)
qflcc.jpg
 
In the first variant, when the dielectric and magnetic constants are considered unchanged, in the reference frame of the remote observer, the Q-factor
the oscillating circuit will be
qflcc1.jpg


 
And in the second variant, when the dielectric and magnetic constants change in different directions by k times, in the reference frame of the remote observer
the Q-factor the oscillating circuit will be
qflcc2.jpg


 
Thus, this thought experiment with an oscillatory circuit shows that in the reference frame of a remote observer, the dielectric and magnetic constants must change in different directions in accordance with the change in the time/distance scale. Then it is not difficult to show
that the Planck constant will also change in the reference frame of the remote observer, since it is equal to
hhhh.jpg

 

For those who still do not believe that the numerical value of the gravitational constant in the reference frame of a remote observer changes back
in proportion to the change in the time/distance scale, I propose to solve the following problem.
On a planet with a mass of 5.97*10 ^ 24 kg, a radius of 6317000 m, an apple falls from a tree from a height of 5 m in 1 second, calculate the value of the gravitational  constant G. And then calculate the value of the gravitational constant G' in the reference frame of the remote observer, in which the apple falls for 1.1 seconds from a height of 5.5 m and the radius of the planet is 6948700, respectively, and compare the obtained values of G and G'.
 
applf.jpg

 

I don't think the acceleration appears to be less to the remote observer, if he is doing his experiment competently.

If he is competent, I think he will calculate the acceleration using a clock on the planet, not one in his lab, in order to correct for the gravitational time dilation effect.   

 

Link to comment
Share on other sites

29 minutes ago, exchemist said:

I don't think the acceleration appears to be less to the remote observer, if he is doing his experiment competently.

If he is competent, I think he will calculate the acceleration using a clock on the planet, not one in his lab, in order to correct for the gravitational time dilation effect.  

Then he should also use a tape measure to measure the distance not from his laboratory. This means that it will switch from a relative coordinate system to an absolute one. In the absolute coordinate system, G is of course a constant.

Link to comment
Share on other sites

3 hours ago, SergUpstart said:

Let another observer from outer space also observe the fall of an apple through a telescope. Due to the fact that the second observer is < br /> in a weaker gravity, in its frame of reference, the time of the apple's fall will be slowed down by a factor of k, and the height of H will also be increased by a factor of k.

By time slowed you mean it takes longer according to that clock, right? The distant observer's clock runs faster than the one next to the tree.

Also, it's not the weaker gravity, it's being higher up in the gravity well that causes time dilation. g could be constant and you would still have this effect.

Link to comment
Share on other sites

6 minutes ago, swansont said:
3 hours ago, SergUpstart said:

Let another observer from outer space also observe the fall of an apple through a telescope. Due to the fact that the second observer is < br /> in a weaker gravity, in its frame of reference, the time of the apple's fall will be slowed down by a factor of k, and the height of H will also be increased by a factor of k.

By time slowed you mean it takes longer according to that clock, right? The distant observer's clock runs faster than the one next to the tree.

Also, it's not the weaker gravity, it's being higher up in the gravity well that causes time dilation. g could be constant and you would still have this effect.

Yes, that's right, by weaker gravity, it meant that the remote observer is higher in the gravity well. But if we substitute time into the equation of motion according to the readings of the remote observer's clock, then the magnitude of the acceleration of gravity g will be different.

Link to comment
Share on other sites

14 minutes ago, SergUpstart said:

Yes, that's right, by weaker gravity, it meant that the remote observer is higher in the gravity well.

Those two conditions are not identical, so one cannot mean the other. 

14 minutes ago, SergUpstart said:

But if we substitute time into the equation of motion according to the readings of the remote observer's clock, then the magnitude of the acceleration of gravity g will be different.

I suspect the problem here is that you are applying SR to a condition where you have an accelerated frame of reference. The issue of length contraction in GR is more complicated, and you are assuming that the transformation between the two frames is the same as in SR. Is that a valid assumption?

 

Link to comment
Share on other sites

11 minutes ago, swansont said:

I suspect the problem here is that you are applying SR to a condition where you have an accelerated frame of reference. The issue of length contraction in GR is more complicated, and you are assuming that the transformation between the two frames is the same as in SR. Is that a valid assumption?

I assume that in order for the speed of light to remain constant, the distance traveled by light and the time interval for which it is traveled would change the same factor.

Link to comment
Share on other sites

22 minutes ago, SergUpstart said:

I assume that in order for the speed of light to remain constant, the distance traveled by light and the time interval for which it is traveled would change the same factor.

We are not measuring the speed of light, or the speed of anything, though. We're measuring an acceleration.

Even in SR, accelerations are not related by a simple multiplication by gamma (or its inverse), so why would you expect this to be true in a calculation of gravitational acceleration? You have naively assumed so, and concluded that G has to change, but I see no basis for this assumption, and no derivation that shows it to be the case.

Link to comment
Share on other sites

18 minutes ago, swansont said:

We are not measuring the speed of light, or the speed of anything, though. We're measuring an acceleration.

Even in SR, accelerations are not related by a simple multiplication by gamma (or its inverse), so why would you expect this to be true in a calculation of gravitational acceleration? You have naively assumed so, and concluded that G has to change, but I see no basis for this assumption, and no derivation that shows it to be the case.

Here we consider acceleration in the non-relativistic case. The apple is falling at a speed much less than the speed of light.

Let's look at the second part of the topic related to the oscillatory circuit. There it is proved that in the reference frame of the remote observer the dielectric constant decreases in proportion to the decrease in the standard length. It is logical to assume that G should change in the same way ( taking into account the fact that the dielectric constant is included in the denominator in Coulomb's law, and G is included in the numerator in Newton's law of gravity )

Edited by SergUpstart
Link to comment
Share on other sites

2 minutes ago, SergUpstart said:

Let's look at the second part of the topic related to the oscillatory circuit. There it is proved that in the reference frame of the remote observer the dielectric constant decreases in proportion to the decrease in the standard length. It is logical to assume that G should change in the same way ( taking into account the fact that the dielectric constant is included in the denominator in Coulomb's law, and G is included in the numerator in Newton's law of gravity )

If you haven't gotten your first example right, how would anyone have any confidence you've gotten anything else correct? Isn't it possible, or perhaps likely, you've made the same mistake?

Link to comment
Share on other sites

5 hours ago, swansont said:

We are not measuring the speed of light, or the speed of anything, though. We're measuring an acceleration.

Even in SR, accelerations are not related by a simple multiplication by gamma (or its inverse), so why would you expect this to be true in a calculation of gravitational acceleration? You have naively assumed so, and concluded that G has to change, but I see no basis for this assumption, and no derivation that shows it to be the case.

Calculating the acceleration of free fall on the surface of the planet is only an intermediate task here. The final task is to calculate the gravitational constant. And in the condition of the problem there are two independent distances, the height of the fall of the apple H and the radius of the planet R. Do you assume that there is such a way to recalculate two independent distances that G will be left behind by a constant?

And it should also be borne in mind that the gravitational well is formed not only (or even not so much) by the gravity of the planet, but also by the gravity of the star, galaxy....

Link to comment
Share on other sites

12 minutes ago, SergUpstart said:

Calculating the acceleration of free fall on the surface of the planet is only an intermediate task here. The final task is to calculate the gravitational constant. And in the condition of the problem there are two independent distances, the height of the fall of the apple H and the radius of the planet R. Do you assume that there is such a way to recalculate two independent distances that G will be left behind by a constant?

You seem to be assuming that the values will have to agree with each other, when we know that instead, such values are relative and can only be compared by transforming from one reference frame into the other.  

I don't know off the top of my head what value a distant observer would calculate for g in a GR context. What I do know is if this were a situation where relative motion were in play, they would not get the same answer for an acceleration, because that's not how accelerations transform, just like they would not get the same answer for kinetic energy of some object in the other frame, and whose value is not found by a simple multiplication by gamma.

Relative values will be relative, not equal, in different frames of reference. Are you arguing that g is an invariant quantity? because that's what you seem to be arguing.

 

Link to comment
Share on other sites

36 minutes ago, swansont said:

Are you arguing that g is an invariant quantity?

I'm just saying the opposite. Quote from the beginning of the topic

"From this it can be seen that in the reference frame of a remote observer, the acceleration of gravity on the surface of planet X should be k times less."

Link to comment
Share on other sites

12 hours ago, SergUpstart said:

the height of H will also be increased by a factor of k.

How come the tree is taller when measured by a remote observer, higher up in the gravity well ? How is that measurement performed? 

I'm curious, I can't remember reading about such an effect (but my GR knowledge is very limited).

Link to comment
Share on other sites

29 minutes ago, Ghideon said:

How come the tree is taller when measured by a remote observer, higher up in the gravity well ? How is that measurement performed? 

I'm curious, I can't remember reading about such an effect (but my GR knowledge is very limited).

If we express the height of the tree in the number of wavelengths of a laser, then it will be the same for all observers. But for a remote observer, this wavelength will be longer due to the gravitational redshift. This means the laser, which is located next to the tree.

Edited by SergUpstart
Link to comment
Share on other sites

5 hours ago, SergUpstart said:

I'm just saying the opposite. Quote from the beginning of the topic

"From this it can be seen that in the reference frame of a remote observer, the acceleration of gravity on the surface of planet X should be k times less."

Noted. But you are insisting that you have to change G, because of this hand-wavy argument, rather than trying to find out the correct transform.

Link to comment
Share on other sites

10 hours ago, SergUpstart said:

If we express the height of the tree in the number of wavelengths of a laser, then it will be the same for all observers. But for a remote observer, this wavelength will be longer due to the gravitational redshift. This means the laser, which is located next to the tree.

I know about the gravitational redshift but I do not see the connection to a change in size. Your statements seem to imply that objects very near the event horizon of a black hole will be extremely long according to an observer far from the black hole. When the redshift is extreme the change in length will be extreme. Is this a correct interpretation of your opening post?

Link to comment
Share on other sites

4 hours ago, Ghideon said:

I know about the gravitational redshift but I do not see the connection to a change in size. Your statements seem to imply that objects very near the event horizon of a black hole will be extremely long according to an observer far from the black hole. When the redshift is extreme the change in length will be extreme. Is this a correct interpretation of your opening post?

No, wrong. The consequence of this post is that the event horizon should not arise at all. If we plot the time dilation function for the mass of a material point

tdgr.jpg.ede347ea0f552d14b7eb5cd7d110bf6c.jpg

The graph of the function T (r)/To looks like this 

graphtd.jpg.d3b58a1db0f0779c273e11948e44ee66.jpg

The graph shows that at r=2GM/c^2, the time dilation becomes infinitely large, i.e. the so-called event horizon takes place.
Now we take into account the change in the gravitational constant inversely proportional to the time dilation, for which we solve the following system of equations. 

 tdgrf.jpg.66adb943e2cda9500b78962fdb7f3f07.jpg

 

The solution of this system of equations

tdmm.jpg.7629e38bf57e694f00e335933d616b9d.jpg

Graph of the received function

tdm.jpg.ef1b00dd30c816959dc4f6ae3e473830.jpg

It can be seen from this graph that the time dilation for an external observer becomes infinite only at r=0.
Thus, there is no event horizon. There are many massive and supermassive bodies in the universe that are asymptotically similar to black holes, but they should not have an event horizon.

 

 

Edited by SergUpstart
Link to comment
Share on other sites

2 hours ago, SergUpstart said:

No, wrong.

Ok! Thanks for clarifying.

2 hours ago, SergUpstart said:

The consequence of this post is that the event horizon should not arise at all.

Do you mean that black holes as described by established theories do not exist? Is your claim that General Relativity is wrong? 

 

Link to comment
Share on other sites

45 minutes ago, Ghideon said:

Do you mean that black holes as described by established theories do not exist? Is your claim that General Relativity is wrong? 

This is not about refuting the GRT, but about correcting it. G in the GRT equation must be a variable whose value is inversely proportional to the time dilation.

The fact that gravity is a curvature of space-time by the momentum energy tensor, the fact that it propagates in the form of gravitational waves at the speed of light, which is constant, all this is beyond doubt.

Edited by SergUpstart
Link to comment
Share on other sites

7 hours ago, SergUpstart said:

G in the GRT equation must be a variable whose value is inversely proportional to the time dilation.

This is meaningless. The GR field equation, being a tensor equation, is a purely local constraint on the metric; thus there’s no time dilation involved.

Besides, if you allow G to vary, the overall equation is no longer covariant, and will have completely different solutions.

11 hours ago, SergUpstart said:

The graph shows that at r=2GM/c^2, the time dilation becomes infinitely large, i.e. the so-called event horizon takes place.

An event horizon is not in general the same as a singularity in your chosen coordinate system. These happen to coincide for Schwarzschild coordinates, but that’s mostly just because of the way these coordinates have been defined. Physically there is no infinite time dilation at the EH - any test particle in free fall will reach the horizon, fall through it, and continue you on down, all in a finite, well defined proper time. It’s just that in-fall time diverges for a stationary far-away observer on his own clock; here again it is crucially important to recognise the difference of global vs local. A Schwarzschild coordinate system is true and valid locally for a stationary distant observer, but tells you nothing about what physically happens at the horizon itself.

Link to comment
Share on other sites

9 hours ago, Markus Hanke said:

Besides, if you allow G to vary, the overall equation is no longer covariant, and will have completely different solutions.

Of course, the solutions will be different and the event horizon and singularity will not appear in them.

Link to comment
Share on other sites

18 hours ago, SergUpstart said:

This is not about refuting the GRT, but about correcting it. G in the GRT equation must be a variable whose value is inversely proportional to the time dilation.

As far as I know the constant G is required in General Relativity to get time dilation and other predictions in the first place. Saying "G in the GRT equation must be a variable whose value is inversely proportional to the time dilation" sounds like a contradiction.

 

18 hours ago, SergUpstart said:

The fact that gravity is a curvature of space-time by the momentum energy tensor, the fact that it propagates in the form of gravitational waves at the speed of light, which is constant, all this is beyond doubt.

Which means that G is constant. 

Link to comment
Share on other sites

17 hours ago, SergUpstart said:

Of course, the solutions will be different and the event horizon and singularity will not appear in them.

Think about this a little more. So you have G=kT, but now allow the constant to become a function (what does it even mean to multiply a non-covariant function with a covariant object such as a tensor? How do you define this consistently?).

First thing you’ll notice is that for T=0 you still get G=0, and thus R=0, which is the exact same as ordinary GR - meaning there’s no change whatsoever to vacuum solutions, including Schwarzschild. You still get event horizons and everything else.

The problems begin in the interior of mass distributions, because now your equation is no longer covariant, meaning the geometry of interior spacetime depends on the observer - which is totally nonsensical, and certainly not what we see in the real world. Furthermore, because of observer dependence you cannot, in general, match the interior metric to the exterior vacuum one while maintaining smoothness and differentiability at the boundary - so it’s not just meaningless, but also mathematically inconsistent. Lastly, the actual interior solutions themselves would now be completely different; using known equations of state for fluids would not result in anything even remotely resembling real-world astrophysical objects such as stars, directly contradicting observation.

So, the idea is nonsensical, mathematically inconsistent, and contradicts real-world observation. And this is just the obvious issues. For example, there’s also deeper reasons why the equations have the specific form they do, but I think the above is already enough to show why this can’t work.
 

16 hours ago, Ghideon said:

Which means that G is constant.

Indeed. And which means you can’t simply change the field equation like that, and expect things to still work.

Edited by Markus Hanke
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.