1. Jun 28, 2008

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.
L: Length of the Pendulum String

Now, let us put the whole experiment (Earth and Pendulum) in a reference frame which is travelling 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.

2. Jun 28, 2008

HallsofIvy

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

3. Jun 28, 2008

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.

4. Jun 29, 2008

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.

5. Jun 29, 2008

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.

Last edited: Jun 29, 2008
6. Jun 29, 2008

yuiop

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.

Last edited: Jun 29, 2008
7. Jun 29, 2008

Shenckel

Dear Kev,

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 travelling 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.

8. Jun 29, 2008

Staff Emeritus
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.

9. Jun 29, 2008

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.

10. Jun 29, 2008

yuiop

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.

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.

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?

11. Jun 29, 2008

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.

12. Jun 29, 2008

Staff Emeritus
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.

13. Jun 30, 2008

Shenckel

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.

14. Jun 30, 2008

yuiop

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.

15. Jun 30, 2008

HallsofIvy

Staff Emeritus
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.

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.

16. Jun 30, 2008

yuiop

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.

17. Jun 30, 2008

yuiop

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

18. Jun 30, 2008

Staff: Mentor

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

19. Jun 30, 2008

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 travelling 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.

20. Jun 30, 2008

JesseM

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.