What is the Classical Pendulum Paradox?

[Shenckel]

Hi,

I have the following problem:

The formula for the Period of a classic pendulum is T=sqrt(L/g)

Where: T: Period of the Pendulum suspended on a string.
g: Earth's acceleration (=G*Mass_of_Earth/(Radius_of_Earth)^2
L: Length of the Pendulum String

Now, let us put the whole experiment (Earth and Pendulum) in a reference frame which is traveling at speed v in a direction perpendicular to the Pendulum string.

The period T of the pendulum should then dilate:

T_newframe = T*GAMMA

Where: GAMMA = 1/sqrt(1-v^2/c^2)
c = speed of light
v = speed of new reference frame

However, looking at the calculated Period of the Pendulum in the new reference Frame, using, the formula for the period, one gets:

T_newframe = sqrt(L/g_newframe) = T/sqrt(GAMMA).

The pendulum length is not changed in the new frame of reference, because it hangs perpendicular to v.

So relativity tells me that the period will become longer because of time dilation, but at the same time it tells me that it will become shorter due to mass increase!

I admit that the Earth will contract into an ellipse or something in the new frame, and calculating g will be a bit more involved, but this does not change the validity of the argument: Instead of putting the pendulum on a spherical Earth, I could have placed it on top of a long, massive bar, which would not change shape due to relativistic effects, and would then give rise to a g_newframe of g_newframe =g*sqrt(GAMMA), as given in the above formula.

Can you tell me where my mistake is?

Thankful for any suggestions,

Sebastian.

 

[HallsofIvy]

For one thing, the angle the pendulum makes changes as it swings so you cannot have a constant velocity perpendicular to the pendulum.

 

[Shenckel]

Let us assume that v_reference >> v_pendulum, and that angle_pendulum<<1 (as is the case with the classical string pendulum). Then all relativistic effects due to the pendulum's angle can be neglected, I presume.

 

[jmnance]

an addition to hallsofivy's comment: time dilation would not behave as it does in normal special relativity because to keep the velocity vector perpendicular to the pendulum bar, one would experience some what of a centripetal force due to the changing direction of the velocity vector and thus be in a non-inertial reference frame.

 

[Shenckel]

I don't really see that the angle the pendulum makes changes anything: I can make the Pendulum swing at an angle as small as I want. Any effect due to the angle of the Pendulum would then become smaller. The velocity of the moving reference frame is assumed to be perpendicular to the 0° position of the pendulum, not to the swinging, i.e. changing, string of the pendulum, so the reference frame is at uniform speed, not accelerated.

 

[yuiop]

...

Can you tell me where my mistake is?

Thankful for any suggestions,

Sebastian.

You are assuming that the gravitational force of the Earth increases with the increase of relativistic mass of the Earth. If you drop an object from a height of 4.9 meters above the ground it takes 1 second to fall. Now if the Earth was moving at 0.8c relative to some observer then the clock of the observer standing on the Earth would slow down by a factor of 0.6 and the object would seem to take 0.6 seconds to fall and he would calculate the acceleration of the Earth gravity to be 27.222 m/s^2 rather than the usual 9.8 m/s^2 when the Earth is at rest.

Obviously, because the laws of physics are the same in all reference frames, that does not happen, so the gravity of the Earth must get less to compensate. That is also the reason why objects moving at close to the speed of light do not turn into black holes.

The pendulum is regulated by gravity. If the force/ acceleration of gravity is less, that compensates for the other factors you mentioned.

P.S. You will not hear this argument from anyone else but it is an unavoidable conclusion of relativity. The subject is generally avoided because it does not sit comfortably with the principle of general relativity that inertial mass and gravitational mass are the same. In a nutshell inertial mass increases with velocity and gravitational mass decreases with velocity, but there are other factors to consider such as the whether you are talking about gravity parallel or transverse to the motion.

 

[Shenckel]

Dear Kev,

thank you for your answer.

To put your example in a formula. For a rock to fall a distance D in a Gravitational Field g, it takes the time:

t=sqrt(2D/g) (Equation 1)

In a reference frame moving at velocity v perpendcular to the Gravitational field g, this drop-time dilates to

t'=t*GAMMA, (Equation 2)

where GAMMA = 1/sqrt(1-(v/c)^2)

So, in order to keep (Equation 1) valid in all frames of reference, the Gravitational field would have to change in the following way:

g'=g/GAMMA^2 (Equation 3)

then, putting (2) and (3) into (1), one gets:

t'=sqrt(2D/g')


Now, if I assume that g is produced by a pointsize mass of Mass M, then the formula for g is:

g = G*M/R^2 (Equation 4)

where:

G: Newton's constant
R: Distance between rock and pointsize Mass (R>>D)

Now,

g' != G*M'/R^2 = G*GAMMA/R^2 = g*GAMMA.

My assumption was that gravity actually seems to become stronger in a traveling frame of reference (not weaker, as you suggest), because the Masses causing it become heavier, and therefore attract more strongly.


I assume my error is in assuming that G'=G, i.e. that Newton's constant is equal in all frames of reference.

If:

G'=G/GAMMA^3, (Equation 5)

then (Equation 3) holds.



So, do you have any reference that can point me to the fact that Newton's 'constant' is contracted by a factor 1/GAMMA^3 when viewed from a moving reference frame?

This would be new to me - I assume it must be an effect of General relativity, which I am not familiar with.

Thanks,

Sebastian.

 

[Vanadium 50]

First, it's not a "paradox" when one is struggling with the calculations.

Second, the treatment of gravitational forces in SR is tricky, which is why we have GR. Strictly speaking you are asking for a relativistic treatment of gravity that is not GR, which is the extension of SR that deals with gravity: i.e. using a theory outside of its range of validity. A mass on a spring would be a better example of a harmonic oscillator.

Third, and perhaps most seriously, you are the victim of the "equation for this, equation for that" mentality. Physics done well starts with the underlying principles. Taking an equation derived under one set of circumstances and applying it in a different set may give the right answer. Or it may not. One of the worst offenders here is "relativistic mass" (which is why the number of the people who work with SR daily and actually use it is approximately zero) - it's introduced to preserve the equation p = mv, but works in no other equation. Certainly just plugging it into Newton's Law of Universal Gravitation doesn't work.

Doing this also causes you to miss some effects - even if you have a flat plane of material, the Lorentz contraction pushes the atoms closer together, increasing the gravitational force. You ignore the Lorentz transformation of acceleration (which is usually not simple), which is a faster way to get where you want to get.

Finding equations that look like they might help and plugging into those is a sure-fire recipe for getting tangled in knots, which is why you are struggling.

 

[Shenckel]

I really believe that Kev is on the right track. Hoping to account for the 'missing gravity' through Lorentz-Contraction effects on the Gravitational mass or the pendulum length is besides the point. The paradox also works with a falling stone - see my previous post.

 

[yuiop]

Second, the treatment of gravitational forces in SR is tricky, which is why we have GR. Strictly speaking you are asking for a relativistic treatment of gravity that is not GR, which is the extension of SR that deals with gravity: i.e. using a theory outside of its range of validity.

Sure, but a basic conclusion of Special Relativity is that it is impossible to detect absolute motion. There is no sub clause that says "unless there is a gravitational field present."
That immediately tells you that the force of gravity of a massive body changes with motion without using the complicated maths of GR. Special Relativity may have difficulty coming up with an exact equation for how much the force and acceleration of gravity changes by, but it does tell you that it certainly does. If anyone here is capable of doing the maths for this in General Relativity they will confirm this, or we will will have to throw out the theory of relativity and embrace aether theory.

... One of the worst offenders here is "relativistic mass" (which is why the number of the people who work with SR daily and actually use it is approximately zero) - it's introduced to preserve the equation p = mv, but works in no other equation. ..

Relativistic mass also appears in the relativistic version of F= ma .. infact it appears wherever we want to analyse the energy equivalent of mass with relative motion or the inertial properties of mass.

This FAQ takes a less negative approach to relativistic mass. http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

"We see that only "γ m" ever appears. The fact that both of these terms appear together is a strong indication that something is going on--that they form a natural pair. This idea of taking our cue from the equations is a powerful tool in mathematical physics."

It is a bit like using t' = t*gamma or L' = L/gamma and denying there is any such thing as time dilation or length contraction.

Finding equations that look like they might help and plugging into those is a sure-fire recipe for getting tangled in knots, which is why you are struggling.

Can you tell us what the force or acceleration of gravity acting on a pendulum with relative motion is, using the equations of General Relativity?

 

[JesseM]

To avoid having to get into general relativity, I would suggest instead making use of the equivalence principle and considering a pendulum or falling stone inside a chamber in space which has rockets that accelerate it at 1G. This still wouldn't be easy to analyze since it involves an accelerating frame, but it's still an SR problem so probably easier than looking at a pendulum in GR.

 

[Vanadium 50]

Sure, but a basic conclusion of Special Relativity is that it is impossible to detect absolute motion. There is no sub clause that says "unless there is a gravitational field present."

Actually, that's exactly what it says. SR says that the laws of physics are the same in all inertial frames. If you want to go to a non-inertial frame, such as one where the bob of a pendulum "spontaneously" accelerates towards the center of the earth, you need GR.

This FAQ takes a less negative approach to relativistic mass. http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

"We see that only "γ m" ever appears. The fact that both of these terms appear together is a strong indication that something is going on--that they form a natural pair.

That's not quite right. Note that there are still gammas left behind next to the velocities. So they don't occur as a pair.

"Relativistic mass" is a term for which we already have a perfectly good name: "energy". It's one component of a 4-vector. (Note that you can always factor out one component of a 4-vector, which is what Baez does. He could just as easily factored out momentum in the x-direction and called in relativistic mass, and he would have gotten a different equation.)

It's a bad idea to give the same thing two different names. It's a bad idea to treat one component of a 4-vector like a scalar. And you most certainly cannot go plugging in one component of a 4-vector into equations meant for scalars (or vice versa) and expect anything correct to pop out.

It is a bit like using t' = t*gamma or L' = L/gamma and denying there is any such thing as time dilation or length contraction. [/B]

No, it's fundamentally different, and I explained why above. This is why of people who use this every day - and I am one of them - that few to none of them use relativistic mass.

 

[Shenckel]

even if you have a flat plane of material, the Lorentz contraction pushes the atoms closer together, increasing the gravitational force.

Vandadium, above you suggest that the decrease in Gravitational force (I suppose you you mistyped "increasing" in the quote above) can be explained by Lorentz-Transformation of the sources of the Gravitational field, i.e. that Lorentz contraction in the Earth-sphere or the bar causing gravitation makes gravitation smaller. This would be wonderful, please show me how this happens, it would make the "conspiracy of relativity", i.e. the phenomenon, that all magnitudes on two sides of an equation change in such a way as to make detection of absolute motion impossible, perfect. However, at this point, I no longer believe this is the cause for the Gravitational force diminishing( but you seem to). Before you discuss any further, you should make your position clear.

As I see things right now, a gravitational field diminishes by a factor 1/gamma^2 in the direction perpendicular to motion. This is something that has to be postulated, and cannot be explained by transformation of the sources, and should be able to be verified by experiment. Even in a time where there was no GR, this fact needed explaining.I think the theme of this thread has changed the subject somewhat.

 

[yuiop]

Actually, that's exactly what it says. SR says that the laws of physics are the same in all inertial frames. If you want to go to a non-inertial frame, such as one where the bob of a pendulum "spontaneously" accelerates towards the center of the earth, you need GR.

The point I think you are missing is that the absolute motion of the Earth is undectable and the fact that the Earth has a gravitational field does not change that. It does not matter whether you are using Special Relativity or General Relativity, the principle underlying all relativity is that absolute motion is impossible to detect.

Put another way: If you can show that General Relativity can determine the absolute motion of the Earth, then you would have a proof of the existence of the aether.

Vandadium, above you suggest that the decrease in Gravitational force (I suppose you you mistyped "increasing" in the quote above) can be explained by Lorentz-Transformation of the sources of the Gravitational field, i.e. that Lorentz contraction in the Earth-sphere or the bar causing gravitation makes gravitation smaller. This would be wonderful, please show me how this happens, it would make the "conspiracy of relativity", i.e. the phenomenon, that all magnitudes on two sides of an equation change in such a way as to make detection of absolute motion impossible, perfect. However, at this point, I no longer believe this is the cause for the Gravitational force diminishing( but you seem to). Before you discuss any further, you should make your position clear.

I think Vandadium is right that length contraction of the gravitational body would by itself increase the gravitational force but there are other factors that "conspire" to reduce the overall transverse force of gravity of a body with relative motion. The Lorentz transformation of transverse force is F' = F/y. This applies to transverse (y direction) gravitational force too (when the source and test particle are co-moving in the x direction).
For example if a mass is suspended from a spring the upward force exerted by the spring is balanced by the force of gravity acting downwards on the mass. When the system is moving the reduction of the upward force of the spring according to the tranverse force Lorentz transformation must be balanced by a corresponding transformation of the gravitational force acting downwards otherwise absolute motion is detectable. I believe just about all the laws of physics can be deduced from the fundamental principle that absolute motion can not be detected. Put another way. If some theory proposes a law of physics that seems to indicate that absolute motion is detectable then that theory is probably wrong.

As I see things right now, a gravitational field diminishes by a factor 1/gamma^2 in the direction perpendicular to motion. This is something that has to be postulated, and cannot be explained by transformation of the sources, and should be able to be verified by experiment. Even in a time where there was no GR, this fact needed explaining.
I think the theme of this thread has changed the subject somewhat.

As mentioned above gravitational force diminishes by 1/y where y=gamma. From the simple F=ma relationship we can deduce that the transformed acceleration of gravity is diminished according to a' = F'/m' = (F/y)/(m*y) = a/y^2.

Going back to the dropped stone thought experiment, it is easy to see that the time for the stone to drop a given height is greater by a factor of gamma according to an observer with motion relative to the Earth system, due to time dilation (t'=t*y) where the unprimed t is the time measured by the observer in the Earth rest frame. For an observer with a velocity of say greater than 0.8c the time dilation due to the gravity of the Earth is insignificant compared to the time dilation due to relative motion. From the acceleration equation d=a/2*t^2 --> a= 2d/t^2 and knowing that transverse distance (d) does not change under transformation then the acceleration according to the observer moving relative to the gravitational system would be a' = 2d/(t')^2 = 2d/(t*y)^2 = a/y^2 which agrees with the transformed acceleration deduced from the Lorentz transformation of transverse force in the previous paragraph.

 

[HallsofIvy]

I don't really see that the angle the pendulum makes changes anything: I can make the Pendulum swing at an angle as small as I want. Any effect due to the angle of the Pendulum would then become smaller.

Yes and eventually you make the angle 0 so the pendulum is not moving at all! I am not talking about the total angle. I am saying that the velocity vector of the pendulum, in its own frame but also relative to any other frame, is constantly changing.

The velocity of the moving reference frame is assumed to be perpendicular to the 0° position of the pendulum, not to the swinging, i.e. changing, string of the pendulum, so the reference frame is at uniform speed, not accelerated.

Then, since length contraction only happens in the direction of the relative motion, you will have to calculate the motion of the pendulum parallel to the motion of the reference frame at each instant- and that is constantly changing.

 

[yuiop]

Hi,

I have the following problem:

The formula for the Period of a classic pendulum is T=sqrt(L/g)

Where: T: Period of the Pendulum suspended on a string.
g: Earth's acceleration (=G*Mass_of_Earth/(Radius_of_Earth)^2
L: Length of the Pendulum String

Now, let us put the whole experiment (Earth and Pendulum) in a reference frame which is traveling at speed v in a direction perpendicular to the Pendulum string.

The period T of the pendulum should then dilate:

T_newframe = T*GAMMA

Where: GAMMA = 1/sqrt(1-v^2/c^2)
c = speed of light
v = speed of new reference frame

However, looking at the calculated Period of the Pendulum in the new reference Frame, using, the formula for the period, one gets:

T_newframe = sqrt(L/g_newframe) = T/sqrt(GAMMA).
Sebastian.

In post #14 I showed some arguments that to a reasonable first aproximation g' = g/(gamma)^2

Using the equation you mentioned for the Period of a classic pendulum T=sqrt(L/g) the transformed period would be T'=sqrt(L/(g/gamma^2) = T*gamma which agrees with the straight forward time dilation transformation of the pedulum period.

Of course if the angle the pendulum swings through is large then there are complications as the string moves parallel to the axis of relative motion and of course the force of gravity changes as the swing gets larger as the the bob moves further away form the gravitational force and no doubt there are other complications but they only serve to provide a smoke screen. To a first order aproximation the force and acceleration of gravity must change with relative motion. If anyone here has the mathematical ability to analyse the full complex situation using General Relativity I would be glad to see it, but I am sure they would arrive at the same conclusion.

P.S. The dropped stone experiment eliminates the angle problem in the pedulum experiment which eliminates one objection.

 

[yuiop]

Yes and eventually you make the angle 0 so the pendulum is not moving at all!

The same argument could be made that in General Relativity the principle of equivalence is only valid in a small volume and is only truly accurate when the volume under consideration is zero. Does that invalidate the principle of equivalence?

 

[Dale]

Can you tell me where my mistake is?

Certainly. It's a pretty common mistake, and I apologize if someone already mentioned it.

in the new reference Frame, using, the formula for the period, one gets:

T_newframe = sqrt(L/g_newframe) = T/sqrt(GAMMA).

The equation you used here was derived for the special case of v=0. It is incorrect to use a special case formula for the general case since often some important terms will have dropped out.

The correct procedure here would be to first derive the equation for a moving pendulum. It should be easy to show that it reduces to the above equation for v=0, and it should correctly "dilate".

 

[Shenckel]

Forget the Pendulum, it is not getting my point across. Kev mentioned the weight suspended on a spring. Let us assume the weight is hovering a wee bit over a bug sitting on the earth. Now we look at the spring-weight-bug-earth system from a reference frame traveling at a high speed vertically to the hanging spring. Both the Masses (weight and earth) will increase by Gamma, their gravitation will therefore increase, the spring will therefore elongate. The bug gets squashed. Or doesn't it?
That is the paradox, in its simplest form.

 

[JesseM]

Forget the Pendulum, it is not getting my point across. Kev mentioned the weight suspended on a spring. Let us assume the weight is hovering a wee bit over a bug sitting on the earth. Now we look at the spring-weight-bug-earth system from a reference frame traveling at a high speed vertically to the hanging spring. Both the Masses (weight and earth) will increase by Gamma, their gravitation will therefore increase, the spring will therefore elongate. The bug gets squashed. Or doesn't it?
That is the paradox, in its simplest form.

Why would you assume that a spring moving at relativistic velocities behaves just like a spring at v=0?

Also, as I said, probably easier to analyze the problem if you make use of the equivalence principle. Imagine an elevator with rockets at the bottom accelerating it at 1G through deep space, and the spring attached to the ceiling with a weight on the bottom will behave just like the same spring in an elevator sitting on Earth's surface. So, you can consider whether the weight is predicted to reach the floor in an inertial frame where this accelerating elevator is temporarily at rest so that the weight's relativistic mass in this frame is equal to its rest mass, and then consider whether the weight is predicted to reach the floor in an inertial frame moving at relativistic speed relative to the elevator, so that the weight's relativistic mass in this frame is significantly greater than its rest mass. Presumably both frames make the same prediction about whether the weight reaches the floor, which means you can't just assume the spring works the same way in both frames and that its amount of stretching depends only on the relativistic mass of the weight attached to it.

 

[Vanadium 50]

Both the Masses (weight and earth) will increase by Gamma

There's your problem right there. You can't plug in [tex]m\gamma[/tex] in Newton's Law of Universal Gravitation and get the right answer. The one and only equation where "relativistic mass" can be plugged in is [tex]p = mv[\tex]. This is one of the reasons why most people who use SR daily have relegated relativistic mass to the junkheap of bad pedagogy.

 

[Shenckel]

Dear Vanadium,

please give me the correct equation for Newton's gravitational law in a moving frame of reference for a point mass. That would be helpful.

Sebastian.

Why would you assume that a spring moving at relativistic velocities behaves just like a spring at v=0?

You're probably right. So let's stick with the falling rock.

 

[Vanadium 50]

please give me the correct equation for Newton's gravitational law in a moving frame of reference for a point mass. That would be helpful.

As mentioned before, this is called "General Relativity". That's your only choice - you can't put an SR bandaid on gravity.

 

[JesseM]

As mentioned before, this is called "General Relativity". That's your only choice - you can't put an SR bandaid on gravity.

If you're just talking about gravity in a small region where tidal forces are negligible, you can just use the equivalence principle and consider and accelerating room in SR, as I said.

 

[Dale]

Forget the Pendulum, it is not getting my point across. ... The bug gets squashed. Or doesn't it?
That is the paradox, in its simplest form.

My comment still stands, and JesseM also made this point. You need to derive the behavior ou a relativistically moving spring from first principles. You will then get an equation which works correctly for all v. You will see that it reduces to the standard equation for v=0 and that it "dilates" correctly. Your mistake is again using the special-case formulas to make conclusions about the general case. That is backwards.

 

[JesseM]

You're probably right. So let's stick with the falling rock.

What is the supposed problem with the falling rock? I'm sure if you have a rock in an accelerating elevator, and clocks on both the wall and the floor of the elevator, and you know the reading on the wall clock as the rock passes it, then all frames will make the same prediction about the reading on the floor clock when the rock reaches it. Of course you'll have to take into account things like the fact that the clocks are ticking at somewhat different rates due to the equivalent of gravitational time dilation which I think is based on the way born rigid acceleration causes different parts of the elevator to have different speeds as seen in an inertial frame.

 

[dmz2]

Hi,

I have the following problem:

The formula for the Period of a classic pendulum is T=sqrt(L/g)

Where: T: Period of the Pendulum suspended on a string.
g: Earth's acceleration (=G*Mass_of_Earth/(Radius_of_Earth)^2
L: Length of the Pendulum String

Now, let us put the whole experiment (Earth and Pendulum) in a reference frame which is traveling at speed v in a direction perpendicular to the Pendulum string.

The period T of the pendulum should then dilate:

T_newframe = T*GAMMA

Where: GAMMA = 1/sqrt(1-v^2/c^2)
c = speed of light
v = speed of new reference frame

However, looking at the calculated Period of the Pendulum in the new reference Frame, using, the formula for the period, one gets:

T_newframe = sqrt(L/g_newframe) = T/sqrt(GAMMA).

The pendulum length is not changed in the new frame of reference, because it hangs perpendicular to v.

So relativity tells me that the period will become longer because of time dilation, but at the same time it tells me that it will become shorter due to mass increase!

I admit that the Earth will contract into an ellipse or something in the new frame, and calculating g will be a bit more involved, but this does not change the validity of the argument: Instead of putting the pendulum on a spherical Earth, I could have placed it on top of a long, massive bar, which would not change shape due to relativistic effects, and would then give rise to a g_newframe of g_newframe =g*sqrt(GAMMA), as given in the above formula.

Can you tell me where my mistake is?

Thankful for any suggestions,

Sebastian.

You can get two types of solutions:

1. In the SR framework, you already know that T'=T*gamma(V) where V is the relative speed of the two frames.

Now, in frame F', L'=L but , when you apply the Lorentz transforms acceleration, you get g'=g*gamma^-2 (this due to the fact that both L and g are transverse to the motion), So, when you calculate T'=sqrt(L'/g') you get gamma* sqrt(L/g)=gamma*T.

So, no paradox.

2. If you want a fancier treatment, Gron has and excellent paper:

Oyvind Gron,
"Covariant formulation of Hooke's Law",
Am. J. Phys. 49 (1981)

 

[Mentz114]

dmz2,
welcome to the forum and thanks for the reference.

 

[dmz2]

dmz2,
welcome to the forum and thanks for the reference.

Thank you.