# Does time dialation cause mass reduction in GR?

1. Jul 27, 2012

### goldk

I love Physics!

I've been thinking about if..... you were in a room with a window, bouncing a ball... and you were looking at another building where you could see a room with a person bouncing a ball, but ..... their time was moving at half speed (Caused by gravitational time dialation only ( not speed)) .... you weigh 170 lbs with respect your scale and they weigh 170 lbs with respect to their scale.

To you.. their ball is also going half speed. ....

Now, somehow you give them your weight scale. Your scale is still on your time though. When they weigh themself your scale would read 85 lbs because the gravity there is only 1/2 the gravity of your time.

My question is.... Is their mass half your mass?

2. Jul 27, 2012

### Staff: Mentor

This is the sticking point in your scenario. How, exactly, are you going to give them your scale while keeping it "on your time"?

3. Jul 28, 2012

### goldk

Thank you for the question. I did say "somehow". It doesn't matter how they would weigh themself on your scale as much as it would be an observation from your perspective. It equally would be hard to set up this scenario in the real world.

Just think of .... from your perception..... they would weigh less, but does that also mean the matter has less mass?

reference...
From Wikipedia... In physics, mass (from Greek μᾶζα "barley cake, lump (of dough)"), more specifically inertial mass, can be defined as a quantitative measure of an object's resistance to acceleration. In addition to this, gravitational mass can be described as a measure of magnitude of the gravitational force which is
1.exerted by an object (active gravitational mass), or
2.experienced by an object (passive gravitational force)

when interacting with a second object. The SI unit of mass is the kilogram (kg).

4. Jul 28, 2012

### Staff: Mentor

Mass is invariant in SR and in GR. If the scale is "somehow" designed such that it reads a smaller number for a moving object, then it isn't measuring mass.

5. Jul 28, 2012

### goldk

How is mass invariant in Special Relativity when it is supposed to grow when the object is moving near the speed of light.

I am really more interested in gravitational time dialation.

In the scenario, I guess you could be in space with a telescope looking down on someone on a planet, bouncing a ball. Their time would be going slower with respect to your time. (For the sake of argument let's just say the gravitational mass is enough where the time on the planet is half the time with respect to your time frame.)

(without getting into large equations) Since it takes twice the time for the ball to reach the ground on the planet; then with respect to you, the ball is half the weight;... is it not?

6. Jul 28, 2012

### Mark M

There is more than one definition of mass. The mass DaleSpam is speaking of is the usual rest mass (or invariant mass) of an object in classical mechanics. This is always agreed upon by all observers. However, there is another type of mass, often called relativistic mass, that is important in special relativity. As you approach the speed of light in some observer's reference frame, this observer will note that you require a much larger force to accelerate. It's given by $\gamma m$. This is what prevents someone from applying a strong enough force to propel you to the speed of light.

No, it isn't. Weight is calculated using the rest mass I mentioned above, which is invariant.

7. Jul 28, 2012

### Staff: Mentor

The invariant mass does not increase when the object is moving near c. What increases without bound is the total energy. Historically, some people doomed the term "relativistic mass" to refer to the total energy divided by c^2, but that is a concept which has largely been abandoned decades ago.

http://en.wikipedia.org/wiki/Invariant_mass

Gravitational time dilation has nothing to do with mass. You cannot just use Newtonian gravitation where gravitational time dilation is significant and expect to get sensible results.

8. Jul 28, 2012

### goldk

What units plug in for γ? I know it is a time factor, but how do you use it?

9. Jul 28, 2012

### Mark M

No, it's the Lorentz factor. It is given by

$\frac {1} {\sqrt {1 - \frac {v^{2}} {c^{2}}}}$

You can multiply by the time measured in some frame of reference to get the corresponding time measured in a reference frame moving with respect to it. Same for length.

EDIT: I can't seem to get the LaTeX working. Just do a google search for the Lorentz factor.

EDIT: Nevermind, fixed it.

Last edited: Jul 28, 2012
10. Jul 28, 2012

### goldk

Thank you! This helps!

11. Jul 28, 2012

### goldk

I think the equation E= γMc^2 can be used for both reference frames as far as velocity goes, however is it not just one way as far as gravitational time dialation goes?

12. Jul 28, 2012

### Mark M

In that equation, γm is the relativistic mass. However, it's just the mass, m, that is relevant to gravity.

13. Jul 28, 2012

### goldk

Cool, Thanks!

14. Jul 30, 2012

### goldk

I just realized that I left out a notable part of my scenario.

"In the scenario, I guess you could be in space with a telescope looking down on someone on a planet, bouncing a ball. Their time would be going slower with respect to your time. (For the sake of argument let's just say the gravitational mass is enough where the time on the planet is half the time with respect to your time frame.)"

If the planet (which has to be a neutron star) has enough gravity to double the time it takes for a ball to reach the ground and come back up...... this is a conundrum.... because with that much gravity wouldn't the ball's bounce period would be so rapid you couldn't even see it bounce? ?????

Time is slower on the star, but the inertial mass seems huge. I'm probably still stuck in neutonian physics.

15. Jul 30, 2012

### Mark M

Firstly, let's ignore gravitational time dilation. Let's say the ball is being dropped on the star, as in your example. Because of the extreme gravitational pull, the ball will reach the ground in a very short period of time. Now, factor in time dilation. The up-and-down bounces can be thought of like the ticks of a clock. So, our observer in space will see them occurring slower than the observer on the star. Since the gravitational pull of the star is very strong, these periods may still be very rapid. But not as rapid as they are measured by the observer on the star.

16. Jul 30, 2012

### Staff: Mentor

Yes, I already mentioned this mistake above. Newtonian gravity is an approximation to GR only in the weak field limit (small mass, low speed, large distances). In any scenario with a significant amount of time dilation you are outside the regime where the approximation is valid. So trying to cobble together time dilation and Newtonian gravity is going to cause problems.

In GR the distant observers observation of a slow fall tells the distant observer something about the spacetime between them and the object, but tells them almost nothing about the mass of the falling object.

17. Jul 30, 2012

### zonde

Can you be more specific? What problems?

18. Jul 31, 2012

### Staff: Mentor

In this case the problem is thinking that the "slower" fall observed due to gravitational time dilation represents some change in mass that needs to be accounted for. That is not the case, the slower fall still follows a geodesic of e.g. a Schwarzschild spacetime with the same mass parameter as detected at a higher altitude.

In other cases that I have dealt with they reached the conclusion that the laws of physics are different in different locations using similar reasoning.

19. Jul 31, 2012

### pervect

Staff Emeritus
The other thing to remember is that different masses fall at the same rate anyway.

So if you drop a feather and a cannonball on the Earth's surface, they'll both accelerate at 9.8 meters/second locally. So you can see that it would be futile to measure the rate at which something falls to determine it's mass.

Yet it seems that was the idea that started the thread as nearly as I can guess - an attempt to guess at something's mass deduced from how fast it falls.

I'm not sure what went wrong where, but clearly something is "funny" with the line of reasoning. Either that, or I've totally misunderstood the point :-(.

This is strictly true only as long as the mass is small enough to be a test mass. Ocasionally someone will mention that dropping even a feather moves the Earth slightly (which is true to some extent, though the details are complicated by the fact that the Earth isn't actually a point body so most calculations of how much it moves are suspect).

20. Jul 31, 2012

### goldk

You all are great! Thank you for your input!